 So good morning, but show today we are going to discuss exercise six of circles. Okay, so I think after this one more exercise is left in our circles. So we will take that also today we will try to complete this exercise six. Okay. So here's our first question. It is saying circles x square plus y square minus two x minus four y equal to zero and x square plus y square minus eight y minus four equal to zero. Few conditions are given in the options. So we have to check which one is the correct. Okay. So let's first write the equation of the circles. It is given that first circle is x square plus y square minus two x minus four y equal to zero. So what is the center of the circle? It will be one comma two. Right. And what will be radius of this circle? It will be one squared one plus two squared four and minus of C. So here it is nothing C is equal to zero here. So our one is coming out to be a root five. Okay. Now let's see the equation of second circle. It is given x square plus y square minus eight y minus four equal to zero. So what is the center of this circle? Center of this circle will be no extrem so x coordinate will be zero and y coordinate will be four. Right. And what is the radius of this circle? It will be zero squared plus four squared sixteen and minus of minus four that is plus four. That is equal to under root of 20. Okay. So under root of 20 we can write it as four into five means two root five. Okay. So if you see we got the center and radius of first circle, we got the centers, coordinates of centers and radius for the second circle. So let's try to find the distance between these centers means C1 C2. What will be the length of C1 C2? So it will be under root one squared one minus zero squared that will be one squared plus two minus four means two squared that is C1 C2 is coming as root of five. Okay. And what will be our R2 minus R1? If you see R2 minus R1 will be two root five and minus of root five. Okay. So R2 minus R1 is also coming as root five. Is it okay? So we got R2 minus R1 as root five and we got this distance between the centers of the circle as root five means both these are equal. So what does it mean? What does it imply? It implies that both the circles will touch each other internally. The diagram if you want to see the diagram for this, it will be something like this. This will be one circle and the next circle will be like this. This will be the situation. So if you see here, let us assume this C1 and C2. So it will be like this C1 C2 will be equal to R2 minus R1. Suppose this is our center one. This is our center two. Okay. So the distance of C1 C2 will be equal to this will be your, this complete thing will be our R1. And this thing will be our R2. Right. So this C1 C2 will be equal to mod of R1 minus R2. Okay. So in this case, our circles touch each other internally. So this option A will be correct. Right. So now let's move to our next question, question number two. The number of common tangents that can be drawn to these circles. Like the equation of two circles are given in our question and the question is asking to find the number of common tangents possible for these two circles. Okay. So let's write the equation of circle. So here is our first circle x square plus y square minus four x minus six y minus three equal to zero. So what is the center of this? What is the center of this circle? It will be two comma three. And what will be radius? It will be four plus nine minus of minus three means plus three. So, well, so it will be four basically. R1 is four and coordinates of this center of the circle is two comma three. Now let's see the other circle. This is x square plus y square plus two x plus two y plus one is equals to zero. Okay. So what will be the coordinates of center? It will be minus one comma minus right. And what will be radius? It will be one plus one minus one means it is radius of the circle is one. Okay. And so now let's find this c1 c2. It will be our length of c1 c2. It will be two minus one minus or means three squared. Three squared plus three minus one means four is squared right. So this c1 c2 is coming out to be 16 plus nine that is five. The root of 25 is five. So this c1 c2 is coming as five right. And our r1 plus r2 if you see our r1 plus r2 is also five four plus four. So what does it mean? It means that both the circles touch each other externally right then only this condition will satisfy. So if the circles touch each other externally then how many tangents can we draw right? How many tangents we can draw if two circles are touches means if two circles touches each other externally. This will be the scenario for this question. So possible tangents will be three like two direct engines we can draw like this one. This will be our one direct tangent. This will be our second direct tangent right and one transverse common tangent can be drawn right. So total number of possible common tangents will be three. How? Two direct common tangents means this is our direct common tangent. Direct common tangent number one. This will be our direct common tangent number two and this will be our transverse common tangent right. So transverse common tangent will be only one. So total number of tangents common tangents will be three okay. And if you see here like how this condition is getting satisfied this will be our C1 C2 right and this is nothing but r1 plus r2. This one is r1 and this distance will be our r2 okay. So whenever C1 C2 is equals to r1 plus r2 it means both touches externally both the circles touches each other externally. I hope this is clear to all. So this is our question number two. Let's see the next question, question number three. It is saying if one of the circles, if one of the circles x square plus y square plus two a x plus c equal to zero and one another circle is there x square plus y square plus two b x plus c equal to zero lies within the other. Then following conditions are given in the option we have to check which condition is getting satisfied means which option is correct for this condition. So it is saying that if you observe it is saying that one circle is totally contained within the other. Right. Suppose this is our one circle. So if you see this circle suppose one more circle is there like this is totally contained in the first circle. Okay, so suppose this one if I'm saying this circle is this circle is x square plus y square plus two a x plus c equals to zero. Okay, and I'm taking this white circle as x square plus y square plus two b x plus c equal to zero. Okay. Now we have to check which option is correct. We can do one thing like. Let me write the common equation of common for both the circles, you will say how common. Yeah, there will be no real common part between these two circles. This common part will be imaginary basically. So, let's try and how to draw the equation of common court. means how to write the equation of common court. It's nothing but our is C1 minus C2 is equals to zero. But let me remind you this common part will be basically imaginary. Why, because these two circles are not cutting each other. There will be no common court in three. It will be imaginary. So this C1 minus equal to zero. So what will be our C1 minus C2, this x square y square will be cancel both these terms are, it will be to a right to to a x minus to be x to a x minus to be x. And C will also be it cancel. So it will be equal to zero. From here, if you see, if I take common to x, then it will be a minus be equal to zero. Okay, from here we can say x equal to zero x equal to zero is it imaginary. x equal to zero is a imaginary common part. And this actually this common card, we can call it as imaginary radical access also in this case in this particularly in this case we can, we can call it as a imaginary radical access also. So x equal to zero. Suppose I'm taking this line as x equal to zero. Yeah, suppose I'm taking this line. Okay, this is as x equal to zero. So this is a equation of our radical access imaginary radical access or imaginary common part. Now, if you observe what is the center of the circle. If you observe what is the center means coordinates of center for this circle it will be basically minus a right minus minus a common zero. Right. And what will be the center for this circle. Suppose I'm telling it as C1 and this as C2. So what will be the center for this, it will be minus B comma zero, right. So x equal to zero is an imaginary common part, but it will not intersect like it will not intersect any of these two circles right x equal to zero this line will not intersect any of these circles. If you observe this minus a means both C1 and C2. Like, if I say this both C1 and C2 lies on same side of x equal to zero right same side of x equal to zero. What does it mean? What does it mean this AB will be always greater than zero. This product of AB will always be greater than zero. And if they lie on the, this right hand side, suppose if they lie on the right hand side of x equal to zero. So this A and B will be the coordinates of center will be positive. What does it mean this A and B both should be negative. If both will be negative the product will be positive. And if they are lying suppose if they are lying this C1 then C2 lies in the left hand side of this x equal to zero, then coordinates of center will be negative. In that case, this A and B both will be positive. So this AB will be greater than zero. And if so is there any options like this? Yeah, we are having AB equal to zero, greater than zero in two of the options. Now we have to check the value of C. So how can we check the value of C? So for checking value of C we can do one thing. Where will the origin will lie? Origin will basically lie on this line, anywhere, anywhere it will lie on this axis. So suppose I am taking this origin as zero comma zero. So this origin is lying outside the circle. We can say this origin is lying outside the, outside the circles, right? And if any point is lying outside the circle, that means if we put that point in the equation of circle, it will be greater than zero, right? So S of origin, S of zero comma zero should be greater than zero. So put zero comma zero in any of the equations it will be zero plus zero plus zero plus C should be greater than zero. From here we can say that C must be greater than zero. So this AB product of AB, product of A and B will always be positive and C will always be positive. So this option third is correct. Option, sorry, option A is correct for this question. Now let's see this next question. The condition that the circle x minus three is whole square plus y minus four whole square is equal to r square lies entirely within the circle x square plus y square is equal to r square. Okay. So let me draw this. So there is one circle, x square plus y square is equal to r square. And there is one another circle, this, sorry, so there is one another circle. Okay, it is lying completely inside this circle. So what is the equation? Suppose I'm taking this circle as x square plus y square is equal to r square, right? And I'm taking this yellow colored circle equation as x minus three whole square plus y minus four whole square is equal to r square. And what is the center of this bigger circle? The center of this bigger circle is basically zero comma zero. And what is the radius? What is the radius for this circle? The radius for this circle is r. Okay. And this is smaller circle if you see. What is the center? What is the center of this? Three comma four, right? Three comma four is the center of this circle and what is the radius? It is r. Okay. So for this circle, this yellow colored circle, to lie completely inside this circle, what will be the condition? The condition will be C1, the distance between C1 and C2 should always be less than r minus r, right? You can put the mod also. So this r minus r should always be greater than C1 and C2. Now, can you find the distance, this C1, C2? It will be three minus zero squared means three squared plus four minus zero squared, four squared under root. So it will be less than our r minus r, modulus of r minus r. Or we can say r minus r modulus should be greater than five, right? This is, this modulus will give always a positive sign only. So we can remove this modulus also and we can simply write it as r minus r is greater than five. Why? Because capital R is greater than a smaller, right? So this will be the condition for this yellow colored circle to lie completely in this white colored circle. Okay. So option B is correct. So this was one question number four. Let's take this next question number five. The circles whose equations are this and this will touch one another externally. Okay. So two circles are given here. And the question is telling that both the circles are touching one another externally if, so we know, we know that if the circles are touching each other externally, it means the distance between their centers must be equal to r1 plus r2 means must be equal to the sum of their radius, right? This is the condition. For touching two circles externally. So let me write the equation of circles. It is given as x square plus y square plus c square minus 2Ax is equal to 0. This is our first circle. And what will be the center for this circle? What will be the center for this circle? It will be a comma 0, right? And what will be the, let me name it as c1. And what will be the radius for this circle? It will be a square plus 0 square will be 0 only. So a square minus c, c means minus of c square, right? Now, let's take the another circle. It is given x square plus y square plus c square minus 2by equal to 0. Okay. So what will be the center of this circle? Center of this circle will be 0 comma because x term is not here. So 0 comma p, right? And what will be the radius of this circle? It will be p square minus of c square. Now apply this condition, apply this condition on this. So c1, c2 means distance between the centers must be equal to r1 plus r2. So what will be our c1, c2 distance between the centers? It will be a square, a minus a square plus b square under root, right? This must be equal to r1 plus r2. That is nothing but a square minus c square plus under root of b square minus c square. Let's square it, square on both sides. So what we will get? Squaring we will get after squaring. We will get a square plus b square is equal to apply a plus b whole square formula here. So it will be a square minus c square. Let me write here clearly. I'm squaring on both ends x, okay? So a square minus c square plus b square minus c square plus 2 under root a square minus c square into b square minus c square, right? So see here this a square and b square will be cancelled out. And this c square after moving to the left hand side, it becomes 2 c square is equal to 2 under root of a square minus c square into b square minus c square, right? So again, this 2 is getting cancelled out. We can do one thing. We can square it once again. After squaring once again, after squaring once again, we will get c to the power 4 is equal to this a square b square minus a square c square minus b square c square plus c to the power 4. Again, it will be cancelled out. And what we will be left with a square b square is equal to a square c square plus b square c square, right? Now divide whole thing by a square b square c square. So if you divide by a square by b square b square c square, so what you will get? What you will get from here? It will be 1 by c square is equal to c square c square will be cancelled means 1 by b square. And here we are left with b square c square will be cancelled 1 by a square. So this is the condition. If this condition is getting satisfied, that is 1 by a square plus 1 by b square is equal to 1 by c square, then both these circles will touch each other externally, right? So this will be our correct option will be option c 1 by a square plus 1 by b square is equal to 1 by c square, right? So this was our question number 5. Let us see this question number 6. So two circles are there with radius r1 and r2, where r1 is greater than r2, greater than or equal to 2. And both these circles are touching each other externally, okay? So if theta be the angle between the direct common tangents, then we have to find the value of theta, okay? So two circles are given here, two circles are given here, right? This one is one circle and we are drawing direct common tangents, right? We have to find the angle between direct common tangents. Now I am drawing one another circle here and what I will do, I will join these points centers of both the circles that will touch, that will pass through that intersection point of both the direct common tangents, right? So suppose this one is our circle with r1 radius, right? This one is our first circle, this one is our circle 1 and this one is our circle 2, right? So if you see, this one is, this is the radius, this will be 90 degree and this is even to be r1. And suppose this is our second circle radius, this is given to be what? r2. So r1 is greater than r2, yeah? So we are drawing in the same way, r1 is greater than r2 and theta is the angle between the direct common tangents. So basically this line joining this c1, c2, this line joining this c1, c2 will intersect, means will pass through the point of intersection of both these direct common tangents. So suppose I am taking this point as q, right? So as per question, this is given that this angle is theta, this angle is theta, angle between both the direct common tangents. Now, if you see, if you see, can we do one construction here? Let me, let me draw, let me do one construction here. What I will do? I will or we can do one thing. Yeah, we can join these points, no? Like, if we join these points, okay, I am joining these points. So this will be basically our r2 minus r1, right? And okay, if we join this also, let me say like, what is going to happen? So what I will do? I will find this theta by 2. So let me take it as alpha. Let me take this angle will be alpha and this angle will be alpha. So basically this alpha plus alpha is equal to theta. So this line c1 q line or c2 q line will basically bisect this angle theta. So 2 alpha is equal to theta. What I will do? I will first try to find this alpha and then we can easily find out this theta also. Multiplying by 2 will give me the value of theta. So what will be alpha basically? If you see, if this angle is alpha, no? So this angle will also be alpha. But this angle will also be alpha, right? If I am not wrong, because these two lines will be parallel basically. These two lines will be parallel. I am drawing this as a rectangle, this r1, r2, r2 and this c2 and c1, okay? So if this angle will be alpha, then we can find this sin alpha as what will be this distance? This distance will be r1 minus r2, right? This distance. So and this will be our, so our sin alpha, this will be our 90 degree. So our sin alpha will become r1 minus r2, right? r1 minus r2 upon this thing. Okay, I have done one mistake basically. These two circles have to be joined now. These are touching each other externally. So this one mistake I made like basically this circle will touch each other and I have given a distance between them. So this is the wrong drawing which I have drawn. So should I erase it and do it again or what to do? Okay, this c1, c2. Let me erase it, right? Or let me draw one more drawing. Let me draw one more drawing. So I am erasing this complete thing. Then that will be better I think. I didn't go through the question. I went on drawing the circles and direct form and tangent. So this was the mistake. But actually both these circles are touching each other externally. So this will be the case. This will be our first circle and this will be our second circle. And now we can draw the direct common tangents. It will be like this. Then drawing one more direct common tangent from this side. So it will be like this. Now it's okay. Joining these two points. I will join the centers to this. Centers to this. Okay. So now what I'm doing, this one is our radius R1. This one is our radius R2. This is this complete angle is theta. This angle will be theta by 2. Okay. So this will be theta by 2. Since this line bisect the angle between the direct common tangents. These are c1, c2, the centers of both the circles. And let me assume this point, the point of intersection of the direct common tangents as q. So what I will do, I will join these two lines. I will join these two lines. And these two. So basically this distance will be, this distance will be R1 minus R2. This distance will be R1 minus R2. Okay. And this, if this angle is theta by 2, this will also be theta by 2. Why? Because these two lines are parallel. These lines and these lines are parallel. Okay. So our sine theta by 2. Our sine theta by 2 will be equal to R1 minus R2 upon c1, c2. And what is c1, c2? c1, c2, the distance between both the centers will be nothing but this one, this one is R1, this one is R2. So this will be R1 plus R2. So what will be our theta by 2? Theta by 2 will be our sine inverse R1 minus R2 upon R1 plus R2. Okay. So, but we need to find that theta. So theta will be nothing but 2 times sine inverse R1 minus R2 upon R1 plus R2. Okay. So this will be our angle between both the direct common tangents. Okay. So this construction was important basically. Once we were drawn this construction, like I have joined this center to this point and I have to join one parallel line to this, like these two lines are parallel and these two lines will be parallel. So this will also be angle theta by 2 1. Right. From here we got this sine theta by 2 value. And once theta by 2 is known to us, we can easily find the theta by multiplying with a factor of 2. Okay. So, let's see this next question. Question number seven, it is saying that the circles this and this intersect each other into distinct points. Okay. So both these circles are intersecting in two distinct points. What does it mean that the distance between them that the distance between them must be less than R1 minus R2. This condition must be satisfied if both the circles are intersecting in two points, two distinct points. Intersecting at two distinct points. Okay. So if we see if you see the equation of trend circles, it is given that x square plus y square minus 10x plus 16 is equals to zero. And our another circle is x square plus y square is equal to R square. So what is the center of the circle? It will be 5 comma zero. Okay. And what is the radius of the circle? It will be 25. It will be 25 minus of 16. That is nothing but three. And what is the center of this circle? It will be zero comma zero and radius is nothing but. Okay. So let's calculate this C1, C2. C1, C2 is easily this five, five only. So C1, C2 is coming to be five. And our, this should be less than, means C1, C2 should be less than what? R1 minus R2 means or more you can say R minus three more. Okay. Or we can say this. No, whether I have applied the, no C1, C2 should be greater than R1 minus R2. For intersecting at two points, let me draw it first. Then it will be easy. So suppose this one is our one circle. This one is our second circle and both these circles are intersecting. Both these circles are intersecting. And this is our C1, C2. So this is our center of the circle. This is the center of the second circle. Right. So if both the circles are intersecting, that means the distance between them should must be less than R1 plus R2. This distance should be less than R1 plus R2. And this C1, C2 must be greater than R1 minus R2. Right. Then only these circles will intersect. Right. So this C1, C2 should be, this will be the condition. This should be the condition for the two circles to intersect at two distinct points. So what I have written here, C1, C2 should be greater than R1 minus R2. This should be greater than, not equal to less than. Right. So this should be greater than, this should be greater than R1 minus R2. This should be greater than R1 minus R2. Similarly, we raise here also. This one will be greater than R1 minus R2. So what does it mean? C1, C2 is 5. Right. 5 should be greater than R minus 3 mod. Okay. So, or we can say R minus 3 mod. Same thing we can write it as R minus 3 mod is less than 5. Okay. So where should this R minus 3 lie? This R minus 3 should lie between 5 to minus 5. Okay. Now add a plus 3. Plus 3 if you add. Where should it lie? Plus 3 means minus 2. R minus 3 and less than, R minus 3 mod is less than 5. No. So R minus 3 should be between 5 to minus 5. Right. So minus 2 and this will be plus 3 if we are adding. It will be minus 2 to 8. And what will be coming from here? R1 plus R2. If you see C1 C2. This C1 C2 should be less than R1 plus R2. R1 plus R2 means 3 plus R mod. Okay. So from here we can say 3 plus R mod is greater than 5. Okay. So either 3 plus R should be greater than 5 or 3 plus R should be greater than 5 or 3 plus R should be less than minus of 5. So from here if you see R should be greater than 2 and R should be less than minus 8. So from this condition we are getting this and this C1 C2 is greater than R1 minus R2. No. So R minus 3. So 5 is greater than R minus 3 mod. Correct. And from here we can say R minus 3 is less than 5. So R minus 3 should lie between this 5 to minus 5. Right. This will be the range of R minus 3. Now if I am adding it minus 3. But what I am getting I don't know why this expression is coming because minus 3. So from here it is coming to be minus 2. The value of R should lie between minus 2 to 8. And from here we are getting R should be greater than 2. R should be greater than 2. And from here we are getting R should be less than 8. So the range of R will be between 2 to 8. 2 to 8. Right. This will be the range of R. This will be the range of R. Have I done some mistake or is it ok? C1 C2 the distance between both the centers is 5 only. Ok. And the radius of circle is 3. The radius of second circle is R. So this C1 C2 should be greater than 3 minus R. Yeah. Right. Or we can say this if you see this if you consider it as 3 minus R. Then this will be within the range of 2 to 8 only. Right. So our answer will be C. Option C. Let's check this with our, what you say, with the diagram once. If I am drawing the in its, this is our circle with center 0 comma 0. Right. And its radius is R. And I am drawing one more circle whose center is how much? Its center is 5 comma 0. Right. So its center is 5 comma 0. And its radius is, this center is 5 comma 0. I am taking second circle. Its radius is 3. So this point will be basically 2 comma 0. Right. This point will be 2 comma 0. This A point will be 2 comma 0. And this B point will be 5 plus 3 means 8 comma 0. Now we have to find this value of R so that it intersect this circle at two given points. So where should R lie? R should be greater than 2 and less than 8. Then only this white colored circle will intersect this circle at two given points. Right. So the value of R from here coming out to be 2 to 8. It should not be more than 8 because if once this R is getting more than 8 means this yellow colored circle will completely contained in that circle. So it will not intersect. So from here also from this graphical approach also we are getting we are getting the value of RS 2 to 8. So option C is correct. So just a minute I am getting one phone call. Yeah. Sorry about you for the interruption. So hope this question is clear to all this question number seven. So we have seen this question both by graphical or geometrical approach and by this inequality approach. Right. So let's move to the next question that is question number eight. So it is saying if the circle X square plus Y square plus 4 X plus 22 Y plus C equal to 0 bisects the circumference of the circle then C plus D is equal to so one circle is there one circle is there which is bisecting the circumference of other other circle. So I've drawn this circle and I'm drawing one another circle. Right. I'm drawing one another circle. Okay. I'm joining these two points, which is the diameter of this. Okay. So as per question, this circle, this X square plus Y square plus 4 X plus 22 Y plus C. This circle if you see the equation of this circle is X square plus Y square plus 4 X plus 22 Y plus C equal to zero. So this circle is bisecting this yellow colored circle right whose equation is the equation of this circle is X square plus Y square minus 2 X plus 8 Y minus D is equals to zero. Okay. So actually what will be the center of this circle, what will be the center of this circle if you see center of this circle will be 2 comma minus 4, 2 comma minus 4, and it will lie on the diameter. In this dia F AB so center will be basically coordinates will be 2 comma minus 4. Since this white colored circle is bisecting this yellow colored circle, what does it mean by setting the circumference means the if we join the point of interaction point of intersection of both the circles that line will pass through the center of the circle and that line will be particularly diameter right this diameter of the circle AB will be diameter of the circle. Since this circle is this white colored circle is bisecting the circumference of this yellow colored circle this AB will be the diameter of circle which will pass definitely through the center of the circle. Okay. So, what we have to find we have to find the value of C plus T. So, if you know, if you know, we can write the equation of common code. We have learnt in our, we have learnt in this chapter we can write the equation of common code AB. This AB is common code to both the circles. However, it will be a diameter for this yellow colored circle. So, equation of the circle is this C minus, suppose I am taking this as circle as C1. Okay, and this circle as C2. So, this C1 minus C2 is equals to 0. Okay. So, this will be x square plus y square minus 2x plus 8y minus d minus of C2. C2 is x square plus y square plus 4x plus 22y plus C is equals to 0. So, this x square, this x square will be cancelled, y square minus y square will be cancelled. Now minus 2x and minus 4x will be minus 6x plus 8y minus 22y. So, this will be minus 14y. And minus d minus C, minus d minus C is equals to 0. Now, this is the equation of AB, which will pass through the center, whose coordinates are 2 comma minus 4. So, let's put it. So, from here, we get minus 6 into 2 will be minus 12, minus 14 into 4, that will be plus 56, right, plus 56 minus, if I take minus common, it will be C plus d is equals to 0. So, from here, if you see center, center will be 1 comma minus 4, right, not 2 comma minus 4, center will be 1 comma minus 4. It's center will be 1 comma minus 4. So, it will be basically minus 6. So, this will be plus 56 minus 6, that will be 50 is equal to C plus d. This is what we need. This is what is asked in the question, the value of C plus d. So, the C plus d will be equal to 50. That means option B is correct. Okay. Now let's take this next question, question number 9. It is saying two circles, this and this are given. Then the equation of circle passing through their point of intersection and the point 1 comma 1. Okay. So, if two circles are given, we can write the equation of other circle means equation we can write the family of equation of family of circles which will pass through the intersection point of these two given circles. Okay. So, our first circle, suppose I'm taking this first circle as x square plus y square is equal to 6, right. And the second circle is x square plus y square minus 6x plus 8 equal to 0. Now the question is asking the equation of circle passing through their point of intersection. So, equation of circle passing through the point of intersection of these two circles can be given as C1 plus lambda times C2 is equals to 0, right. So, let's put it. So, this will be x square plus y square minus 6, right, plus lambda times C2. C2 is nothing but x square plus y square minus 6x plus 8 is equals to 0. Okay. And this circle, this circle is also passing through this point 1 comma 1. Okay. So, first, make it simpler like this will be 1 plus lambda x square, right, then 1 plus lambda y square, then it will be minus 6 minus 6 from here minus 6 lambda x plus 8 lambda right, this will be equal to 0. So, now put this point 1 comma 1 here 1 comma 1 in this equation, right. So, putting this 1 comma 1 in this equation what we will get, we will get 1 plus lambda plus 1 plus lambda into 1 minus 6 minus 6 lambda plus 8 lambda is equals to 0. So, what it will become lambda plus lambda 2 lambda 2 lambda plus 8 lambda 10 lambda minus 6 lambda will be 4 lambda and 1 plus 12 2 minus 4 minus 5 is equals to 0 from here we got lambda as 1. So, we can write the equation of the required circle as this x square plus y square minus 6 plus lambda is equal to 1 so it will be x square plus y square minus 6x plus 8 is equals to 0. So, x square plus x square is 2x square, okay plus 2y square then we will have minus 6x and plus 8 minus 6 that is plus 2 is equals to 0. You can make the coefficients of x square and y square as 1. So, divide this whole thing by 2 we will get x square plus y square minus 3x plus 1 is equals to 0. This will be our equation of required circle which will pass through the intersection point of this c 1 c 2 and that circle will also pass through point 1 comma 1. So, x square plus y square minus 3x plus 1 equal to 0 option piece there. So, it will be correct option. Let's see the next question, question number 10. The equation of circle described on the common chord of circles this and this as diameter. Okay, means it is saying that this question is saying that two circles are there, okay. They are intersecting also. So, these are two circles. Equation of circle described on the common chord. Okay. So, we have to write one equation of circle which taking this as chord. So, here it is one circle is given. This one is our second circle and the common chord. There will be one common chord also, right. This will be our common chord. Common chord means join the point of intersection of both these circles. This will be our common chord. Now the question is asking to write the equation of circle having this common chord as diameter means having this a b as diameter we have to write the equation of the circle. Okay, suppose I'm taking this circle as C1, C2. It is given in the question itself. So, we have to write the equation of that circle whose diameter is a b. Okay. So, here it is given. First circle is given as x square plus y square plus 2x is equals to 0. Okay. And the second circle is x square plus y square plus 2y is equals to 0. Okay. So, let me first write the equation of a b. Let me first write the equation of a b. It will be C1 minus A2 equal to 0. Okay. So, it will become x square plus y square plus 2x minus x square minus y square minus 2y equal to 0. So, x square minus x square y square minus y square it will become 2 times x minus y equal to 0 or x minus y equal to 0. So, we got the equation of a b. Right. We got the equation of a b. Now, making this a b as diameter, we have to write the equation of that circle. We have to write the equation of that circle. So, we can do one thing. We can find the coordinates of this a and b. Okay. If we can find the coordinates of this a and b, then we can write the equation of the required circle taking a and b as the diameter equals. Okay. So, let's try to figure it out. Let's try to find the coordinates of a and b. So, this x minus y equal to 0. This one is one equation. And let me suppose take C1. I will, what I am trying to do, what I will solve for x and y. So, this x square plus y square plus 2x equal to 0. I will take the intersection problem. So, from equation 1, y equal to x. So, I can put here y square plus y square plus 2y equal to 0. Okay. From here, we get 2y square plus 2y. This will be, okay. Let me write it completely. 2y square plus 2y equals to 0. So, what I will be having y plus 1 equal to 0. Okay. So, let's go solve kj. Now, y equal to x. No, we have put y in terms of x here. So, we got this quadratic. From here, we get y equal to 0 or y equal to minus 1, y equal to 0 or y equal to minus 1. Okay. And since x is equal to y, our x will be equal to 0 and it will be equal to minus 1. Okay. Similarly, you can find the point of interaction by taking this circle C2 also. So, the coordinates of A and B we got as x is 0. Then y will also be 0. And if x is minus 1, y will also be minus 1. Right. So, now we got the A, B coordinates of A and B. So, our required circle will be, our required circle will be x minus x1 into x minus x2 plus y minus y1 into y minus y2 equal to 0. So, this one x, suppose this one is your x1, y1, right? And this one is your x2, y2. So, it will become x into x minus 1 means x plus 1 plus y into y minus y2 will be y plus 1 equal to 0. So, it will become x square plus x plus y square plus y equal to 0. Or we can simply write it as x square plus y square plus x plus y equal to 0. This is our required circle. Suppose, which is passing through the intersection, sorry, which is having A, B as its diameter, right? x square plus y square plus x plus y equal to 0. So, this option D is correct. Now, let's take this next question, question number 11. The equation of the diameter of the circle this, which is perpendicular to the line 2x plus 3y equals to 12. Okay. So, one equation, one circle is there. And the question is asking to write the equation of the diameter of this circle, which is perpendicular to line this, means which is perpendicular to, so suppose I am taking this as diameter and that will, this line will be perpendicular to this line. This is what the question is saying. So, this line is given as 2x plus 3y is equals to 12. Okay. Or we can simply write it as 3y is equals to minus 2x plus 12. So, this will be our diameter. Suppose this is our A, B, right? So, A, B will be the diameter. And this diameter is perpendicular to this line. This is what is saying in the question. So, we have to find the equation of this A, B. Okay. So, let me write the equation of the circle first. It is 3 times x square plus y square minus 2x plus 6y minus 9 equals to 0. So, divide it by 3. We get x square plus y square minus 2y 3x, right? Plus 2y minus 3 equal to 0. So, what will be the center for this? What will be the center for this circle? It will be 1 comma 3 comma, 1 comma 3 comma minus, right? This will be the center of this circle. 1 comma 3, sorry, 1 by 3 comma minus 1. Okay. So, we can assume the equation of A, B, right? Which is passing through this point. So, y minus y1, y minus y1 is equal to mx minus x1. Now, what will be the slope for this A, B? Slope of A, B. Suppose this is, I am taking a slope of A, B. Into slope of, suppose I am taking this line as CD. Slope of CD will be minus 1, since both these lines are perpendicular. So, we need to find the slope of A, B. Slope of A, B, I am taking as m. And what will be slope of CD? It will be from here, if you see, y equal to minus 2 by 3x plus 4. This is what this equation is becoming. Dividing by 3 minus 2 by 3 into x plus 4. So, slope of CD is minus 2 by 3 minus 2 by 3 and that should be equal to minus 1. So, our slope of A, B is coming out to be 3 by 2. So, we are done with this question. We will be able to easily find out the equation of circle y plus 1 is equal to m is 3 by 2. And what is this? 3, then 3x minus 1. So, this 3 and 3 will be cancelled out. This will be 2y plus 2 is equals to 3x minus 1. So, 2y minus 3x 2y minus 3x plus 3 is equals to 0. So, what we can do? We can take a negative common minus sign common. So, this will be 3x minus 2y minus 3 equals to 0. 3x minus 2y minus 3 equals to 0. So, option B is correct. Okay. Hope the calculation part is okay here. So, 3, 3x minus 1. m is slope of A, B is coming to be 3 by 2, right? So, 2y plus 2 equal to 3x minus 1. Yes, it's okay. So, any equation can be either this or this. So, as per option R, B option is coming 3x minus 2y minus 3 equal to 0. And this question is done question number 11. Now, we are going to start this question number 12. So, if the curves this and this intersect at 4 concitalic points, then the value of A. Okay, if you observe the equation, this is not a equation of a circle basically because x term is 0. So, anyway, if two curves are intersecting, no, if two curves are intersecting, we can write the equation of curve passing through the intersection point of these two curves as s1 means c1 plus lambda times c2 is equals to 0 where c1 and c2 are any curves. c1 and c2 are any two curves. So, we can write this as c1 plus lambda times c2 is equals to 0. Okay, this curve will pass through the intersection point of this c1 and c2. Is it okay? So, as per question, it is saying that they are intersecting at 4 concitalic points. So, let me first simplify it and write it. So, c1 is nothing but ax square plus 4xy plus 2y square plus x plus y plus 5, right, plus lambda times, this one is our c1 plus lambda times c2. c2 is ax square plus 6xy plus 5y square plus 2x plus 3y plus 8, right, this should be equal to 0. Now, it is saying that this curve is, means these two curves are intersecting at 4 concitalic points. What does it mean? All the four points lie on the circle. So, this resulting curve, this resulting curve should be the equation of circle. Then only they will fall on the concitalic points, they will lie on the concitalic points. So, this curve should be circle basically. This curve should be circle. Okay. So, let's simplify this curve. What we can write? We can write a plus lambda x square, right. Then, if you take xy term, it will be 4 plus, xy means from here it will be 6 lambda, 4 plus 6 lambda into xy. Now, take y square terms, it will be 2 plus 5 lambda y square, right. Then, x plus lambda 2, x plus 2 lambda, 1 plus 2 lambda into x, then 1 plus 3 lambda into y. Okay. And then 5 plus 8 lambda, 5 plus 8 lambda is equal to 0. This curve should be circle, okay. So, we know for circle, this xy term, this 4 plus 6 lambda term should be equal to 0 and coefficient of x square, coefficient of x square should be equal to coefficient of y square, right. This is the condition for a general second degree equation to be a circle, okay. From here we get 6 lambda is equals to minus 4 or lambda is equals to minus 2 upon 3, right. And what is the coefficient of x square here? It is a plus lambda and what is the coefficient of y square here? 2 plus 5 lambda, right. 2 plus 5 lambda. So, and we need to find the value of a. So, a plus lambda is how much? Minus 2 by 3. So, minus 2 a by 3 is equals to 2 plus 5 into lambda is minus 2 by 3. So, if you take ncm, we get 3 a minus 2 a is equals to how much it will be coming? It will be 3. So, 6 and minus 10, right. So, from here if you get a will be equal to 3 a minus a 2 a, a will be equal to minus 2. So, this will be our answer. This will be our answer. Now, how, why we said that this curve should be circle? That is important thing in this question because it is saying that these two curves intersect at four concyclic points, right. So, four concyclic points means the resulting curve should be a circle. So, a is coming out to be minus 4. So, option b is correct. Option b is correct. Now, let's take the next question, question number 30. Find the equation of the circle passing through the point of intersection of this and this and point 1 comma 1. So, we have done this type of question earlier in this exercise. So, our resulting circle, our required circle will be of, our required circle will be of forms c1 plus lambda times c2 is equals to 0. What is c1? It is x square plus y square plus 13x minus 3y, right. Plus lambda times c2, c2 is 2x square plus 2y square plus 2y square plus 4x minus 7y minus 25 equal to 0, right. But there is one catch. We can, we have to make this coefficient of x square and y square as 1, right. So, I have to make the coefficient of this x square and y square as 1. So, I have to divide this whole thing by 2, right. So, it will become x square plus y square plus 13x minus 3y plus lambda into x square plus y square plus 2x minus 7 by 2y and minus 25 by 2 is equals to 0. Now, this circle is also passing through 1 comma 1. So, it should satisfy this 1 comma 1. So, this will be 1 plus 1 plus 13 minus 3 plus lambda into 1 plus 1 plus 2 minus 7 by 2 minus 25 by 2 is equals to 0, right. So, 1 plus 1, 2, 13, 13 plus 15 minus 3, 15 minus 3 will be 12, 12 plus lambda, 1 plus 1, 2, 2 plus 2, 4, 4 minus this, let me take LCM completely. So, 2, then this will be 4, 2, 4, 2, 8 minus 7 minus 25, right. This should be equal to 0. So, 12 plus, how much it will be? 8 minus 7 will be 1, 1 minus 24 means minus 24, minus 24 by 2, that will be minus 12. So, 12 plus minus 12 lambda is equals to 0. So, from here we got lambda is 1, right. 12 lambda is equals to 12. So, lambda is equals to 1. So, what will be our required equation of circle? From our required equation of circle will be. So, put lambda is equals to 1. So, it will be x square, x square plus y square plus 13x minus 3y, right, plus x square plus y square plus 2x minus 7 by 2y and plus minus 12. So, after simplification what we get? 2x square plus 2y square plus 13x plus 2x, that is plus 15x and minus 3y minus 7 by 2y, that will be 3 and 36, 6.5 minutes minus 13 by 2y. And minus 25 by 2 is equals to 0, right. Further, we can divide it by 2 to make the coefficient of x square and y square as 1, but this will be also correct answer, not an issue. Further, you can simplify and you can make it any like after dividing 2, you can make it as another equation making the coefficient of x square and y square as that is also valid. So, hope this is clear to all. 2x square, 2y square, 13x plus 2x, that will be 15x and minus 13 by 2y and minus 25 by 2 equal to the second. So, this will be our required equation of circle. Now, it is saying that so that the common chord of the circles this and this pass through the center of the second circle and find its length, pass through the center of the second circle, right. So, let me draw it, let me draw another circle. It is saying that so that the common chord of the circle this and this pass through the center of the second circle, okay. So, like if I join the common point, next point of intersection, okay. So, it is saying the common part of the circle means this white colored circle I am taking this circle as I am taking this circle as x square plus y square minus 6x minus 4y plus 9 equal to 0, right. And this circle is the equation of this circle is x square plus y square minus 8x minus 6y plus 23 equals to 0. So, it is saying that suppose this is our circle 1, this is our circle 2. So, this circle 1 and circle 2 are intersecting. So, this A and B will be the common chord for both the circles. Now, it is saying that so that this AB will pass through the center of the second circle and find its length. So, if you see, if you see what will be the center of this circle, what will be the center of this circle, the center of this circle will be 4 comma 3, right, 4 comma 3. And what will be the radius of this circle? It will be 16 plus 9 and minus 23, that is 16 plus 9 is 25, okay, 25 minus 23 is 2. So, R2 is coming to be root 2. This will be the radius of this circle. So, let me mark it as this A2, the coordinates of this will be 4 comma 3, right, just a minute, just a minute. So, this, the center of this circle will be 4 comma 3 and its radius is root 2. So, let me write the equation of AB, equation of AB. How can we write the equation of AB? C1 minus A2 is equals to 0. C1 minus A2 equals to 0, where C1 is this x square plus y square minus 6x. So, this will be minus 6x plus 8x, that will be 2x, right, 2x minus 4y and then plus 6y. So, this will be plus 2y. So, and plus 9 minus 23, minus 23 plus 9, it will be 14. Okay. Now, let's put or we can simply write it as x plus y minus 7 is equals to 0. Now, let's put this coordinates of C2 here, coordinates of C2 here. Let's see whether it is satisfying or not. So, coordinates of C2 is 4 comma 3. So, let's put here. So, it will be 4 plus 3, 7 minus 7 is equals to 0. Yeah. So, it means this line is passing through. Since it is satisfying, it passes through C2. That has been through. It passes through C2. Okay. Now, if it passes through C2, what does it mean? It means this AB must be the diameter of C2. It means AB must be the diameter, must be dia of circle 2. And if it will be the diameter of circle 2, what will be the length of AB? It will be 2 times r. That is nothing but 2 root 2. Okay. So, this will be our answer. We have proved that this line will pass through the center and we have also proved that. We have also proved that its length is means we have also find out its length as 2 root 2. Okay. So, this is what it is asking the question. So, let's take the next question, question number 15. It is saying prove that the circles this and this touch each other. If AB1 is equals to A1B. Okay. For touching the circles, like 2 circles are here, x square plus y square plus 2A is plus 2B y equal to 0. This one is our mean circle. And our second circle is x square plus y square plus 2A1x plus 2B1y equal to 0. Okay. So, what is the center for this? What is the center for this? This will be minus A rate minus A comma minus B. Right. And what will be the center of this circle? It will be minus of A1 comma minus of B1. And what will be radius of this circle? It will be A square plus B square. No constant volume. So, it will be under root of A square plus B square. And here it will be under root of A1 square plus B1 square. Okay. Now, the question is asking that prove that these 2 circles will touch each other if this is AB1 is equal to A1B. So, what is the condition for 2 circles to touch? They can either touch externally or they can touch internally also. So, this C1 C2, this C1 C2 should be equal to if R1 plus R2 means they are touching externally. And if R1 minus R2, if this will be in the distance between both the centers will be R1 minus R2, it means both will touch internally. Right. So, this is the condition for touching. To touch. This is the condition for touching 2 circles. Condition for touching 2 circles. Right. Now, what is C1 C2? C1 C2 will be minus A minus B means A1 minus A whole square plus B1 minus B1 minus B whole square under root. And this must be equal to R1 plus minus R2. That means A under root of A square plus B square plus minus under root of A1 square plus B1 square. Okay. Now, we square it on both sides. What we will have? After the squaring, we will have A1 square plus A square minus 2AA1, okay, plus B1 square plus B square minus 2BB1 is equals to A square plus B square plus A1 square plus B1 square plus minus 2 under root A square plus B square into A1 square plus B1 square under root. Right. So, this A square, A square will be cancelled out. This B square, B square will be cancelled out. And we are left with this. Okay. This will also cancelled out. A1 square, A1 square, B1 square, B1 square. So, we are left with minus 2AA1, okay, and minus 2BB1 is equals to plus minus 2 under root of this. A square plus B square under root of A1 square plus B1 square. Okay. So, let's take this negative common. So, this sign will be plus. Okay. And this 2 will also be cancelled out. This 2 we can also cancel it out. So, and again squaring if you square it again, what you will get if you square it again, you will get AAA1 plus BB1 square. Okay. A is equals to, this square root will be removed, means A square plus B square into A1 square plus B1 square. So, this will be A square A1 square plus B square B1 square plus 2AB A1B1, 2AB A1B1 is equals to A square A1 square plus A square B1 square plus B square A1 square plus B square B1 square. Okay. So, A square A1 square, A square A1 square will be cancelled. B square B1 square will be cancelled. Now, we are left with this one. 2AA1 AB1 and 2AB1 and BA1 is equals to this A square B1 square plus B square A1 square. Now, as per condition, if A1 AB1 is equals to, this is what we've got if the circles are touching each other. So, obviously, if AB1 is equal to, like, if I take this AB1 is equal to, if AB1 is equals to A1B, right? So, what we can write here? We can write here two times or let me take these negatives left-hand side as it is. What we can write this? This RHS will become what? This will be AB1 squared, no? So, I'm writing, in place of this, I'm writing this as A1B squared, this first term. A1B squared and this will be as it is. A1B squared. A1B squared. So, this is nothing but two times A1 squared B squared. Or A1 squared B squared, no? So, A1 squared B squared. So, our RHS becomes this and if you take this AB1 as A1B. So, it will become AB1 means we are replacing AB1 by A1B. So, it will become two A1B into B. Okay? So, similarly, here also, in left-hand side, we are getting two B squared into A1 squared and which is equal to right-hand side. So, yeah, if this condition is there, then both the circles will touch each other. Now, let's move to the next question. This is the last question. I don't know what is the length of this video, but we will close this question as soon as we can. I will not solve the complete question this, this 16th one, but I will give the idea like what is to be done here. So, find the equation of common tensions to this and this. Okay? So, let me first write the equation. So, find the equation of common tension. So, two circles are given here and we have to find the equation of common tensions. So, first we have to identify the position of circles like whether they are intersecting or they are contained in one another or they are separated. Let's see what is here. So, this will be x square plus y square minus 24x plus 2y plus 120 equal to 0. This is our one circle. Now, next circle is x square plus y square plus 20x minus 6y minus 116 is equals to 0. So, what is the center of this circle? It is 12 comma minus 1. And what is the radius of this? The radius of this will be 144 plus 1 minus 120. That is 145 minus 120, that is 5. And what is the center of this circle? It will be minus 10 comma 3. And what will be the radius? The radius will be 100 plus 9 minus 116 means plus of 116. So, 100, 200, 225. 225 means R2 is equals to 15. So, let me check what is this C1 circle? What is the distance between the centers? So, it will be 12 minus 10 means 22 is squared. 22 is squared and plus 4 is squared. So, 22 into 22 will be 22 into 22. What it will be? This will be 44. 44 and 44. This will be 484 plus 16. 484 plus 16 means 500 under root of 500. So, this will be 100 into 5 means 10 root 5. This will be greater than R1 plus R2. Means R1 plus R2 if you see, R1 plus R2 is to be 15 plus 5 means 20. And this 10 root 5, root 5 means it will be greater than 2. So, this will definitely greater than 20. So, means what? Both the circle just mute. It's mean both the circles are separated, right? They are not intersecting. And in this case, total four tangents will come. Let me draw it like how these circles will look like. So, this will be our bigger circle. Okay. So, we'll draw over the tangents here. The other circle will be smaller one. So, this will be our smaller circle. And let me join these three points. These three points will lie on the same circle. So, these, the two direct common tangents. And two indirect common tangents will be there. Four tangents will be there. This will be your, this will be our direct common tangent one, direct common tangent one. This will be our direct common tangent two. 2. This will be our transverse common tangent 1 and this will be our transverse common tangent 2, right? Suppose this is our C2, C2 is having bigger radius, okay? So this C2 will be minus 10,3 minus 10,3. This will be our C1 which is 12 comma minus 1, 12 comma minus 1, the coordinates of both the centers. This is the radius, right? Its radius is 15, its radius is how much? Its radius is 5 and let me take these points as p and let me take these points as q. Means p where the transverse common tangents are intersecting and q is the point where both the direct common tangents are intersecting. So one condition will means you must know one thing that C2p, that C2p, this C2p upon the C1p, C1p, this should be in the ratio of their radius that is 15 by 5 that is 3 is to 1. So p, p divides internally, p divides, p divides this C1, C2 internally, internally in ratio 3 is to 1, right? In ratio 3 is to 1. This you must know, okay? And if that is case, we can anyhow find the coordinates of p since the coordinates of C1 and C2 are unknown. So once this coordinate is available with us, okay? So we can write the equation of line passing through p, okay? So one more condition we need for that. What we need? We need the tangent, sorry, we need the slope of this line, TCT. So for slope, if you see, you can drop one perpendicular from this C2 on this line, right? You can drop one perpendicular from C2 on this line or from C1 on this line, okay? This will be equal to the radius, right? So from there, you can find the equation of TCT. And for direct common tangents, if you observe this q, this q will divide, this q will divide this C1, C2 externally, externally in the ratio 3 is to 1. So this thing you should know so that the coordinates of p and q can be identified. Once we know the coordinates of p and q, we write the equation of the direct common tangent and indirect common tangent assuming y equal to mx plus C or y minus y1 is equal to mx minus x1, where x1 and y1 are the coordinates of p and q. And for m, what we will do? We will drop a perpendicular from the centers on these tangents and we will equate that distance equal to the radius. From there, we will have the two values of m, which we will put in the equation and we will find the value of, I mean, the equation of the tangents. So this is all for today. We have done this exercise 6 of circles. So I think one more exercise is left in this chapter. We will cover that also very soon. So till then, okay. Tata, goodbye. Okay.