 So, good morning everyone. Today, we are going to discuss exercise four of this circle. So, let's see the first question. It is saying that the length of car cut off by y equal to 2x plus one from the circle x square plus y square is equal to two. Let me see whether I have started the recording or not. Yeah, it is going on. So, yeah, let me draw the sketch for this. So, it is saying one circle is there having center as origin, right. So, this is our circle having origin as the center and one straight line is there equation of a straight line is given. So, this is our straight line and it is saying what will be the length of car cut off by this straight line on the circle. Okay, so let me do one construction here. I am joining this center to this point and I'm dropping one perpendicular on this line from the center. So, now we are done with the sketching portion. So, let me name it. So, this line is 2x minus y plus one is equals to zero, right. This is our center of the circle having coordinates as zero comma zero and this is the intercept length of intercept. This is what is asked. The length of AB is asked in the question, right. So, from the center what I have done, I have dropped one perpendicular on this straight line. Let me call it as D. So, if you see in the triangle ADC, we have AD square plus CD square is equal to AC square. Okay, and what is the equation of this circle, if you see this is nothing but x square plus y square is equals to two. So, our AD square will become this AC square minus CD square, right. Now, what is AC? AC is the radius of the circle. AC is the radius of the circle. So, AC square is known to us. We need to find this CD. So, what will be CD? CD is the perpendicular distance from center on this line. So, it will be zero plus zero means two into zero minus one into zero plus one upon under root of A square plus B square that is two square plus one. This will be one upon rule five. So, CD square will be one upon five. So, this AC square is nothing but R square which is two minus CD square is one upon five. Sorry. So, this will become 10 minus one nine upon five. Okay, so AD square we got as nine by five. So, what will be AD? AD will be equal to three upon root five. And what will be AB? AB is nothing but two times AD, right, because this CD will bisect this chord AB. So, this will be two times AD that will be nothing but six upon root five, right. So, option C is correct for this one. So, now see this question number two. It is saying that the circle X square plus Y square minus four X minus H Y minus five equal to zero will intersect the line in two distinct points if okay. So, you know that if a straight line, like, suppose this is any circle. And one straight line is there which is intersecting the circle at two distinct points. It means that the perpendicular distance from the center of this circle on this line should be less than radius of the circle, right. So, this is what we need to do here. So, this is the equation of circle X square plus Y square minus four X minus eight Y minus five is equals to zero. So, what is the center of the circle? The center of the circle is two comma four. And what is the radius? It is two is squared plus four is squared minus of C that is minus of minus five four plus five. Four plus 16 plus five under root 25 that is nothing but five. So, we got the center of this circle as two comma four and we got the radius as five. And this is our line given line whose equation is three X minus four Y plus lambda is equals to zero. Okay, so what I will do, I will drop one perpendicular from C. I will drop one perpendicular from C on this line, okay. And I will name it as D and let the intersection point be A and B, okay. So, for this straight line to have two intersection points on this circle, we need CD must be less than the radius of the circle which is nothing but five. So, this CD must be less than five. And what will be CD? So, our perpendicular distance from center on this line will be three into two that is six minus four into minus four minus 16 minus lambda, minus lambda, sorry. Mod upon under root of A square plus B square that is three square plus four square this must be less than five. So, this will be the equation, the equation of line is given as three X minus four Y minus of lambda, right. This will be minus of lambda. So, three X minus four Y minus lambda equal to zero. So, this will be the perpendicular distance on this straight line from the center and that must be less than five, right. Now, calculate it, this will be minus 10 minus lambda come out should be less than this five into five, 25, okay. So, if you know if mod X is less than eight, so X must lie between eight and minus eight, right. So, we can from here we can write this minus 10 minus lambda. This must be between minus 25 to 25, okay. So, let's add plus 10. So, it will be minus 15. This will be minus 25 plus 10 that will be minus 15 minus lambda. So, multiply by minus one. So, the sign of inequality will change. So, this is 15 is greater than lambda, greater than minus of 35, right. So, lambda should lie between minus 35 to 15. So, lambda should lie between minus 35 to 15, okay. So, this will be our answer, yet option C is correct. Now, see the next question, question number three. It is saying if the line three X minus four Y plus lambda equal to zero where lambda is positive touches the circle at A and B, okay. The value of lambda plus A plus B is equal to, okay. So, the given line touches the circle. It means the given line is a tangent to the circle, okay. So, let me draw it. This is our circle and one straight line is given three X minus four Y. This is our straight line three X minus four Y plus lambda equal to zero touches at A and B, okay. So, our circle is X square plus Y square minus four X minus eight Y minus five equal to zero, okay. And the equation of this line is three X minus four Y plus lambda is equal to zero. And this is a, this is touching the circle. Suppose I'm taking it as B, the coordinates of those are given as A comma B. This is what is given. This line touches the circle at A comma B, okay. And what is the center of this circle? The center of this circle will be two comma four. And what is the radius? Radius will be two is squared four plus four is squared sixteen minus of minus five means plus five. Radius is five. So, this is what we got from this circle. The center is two comma four. And its radius is five, okay. So, the distance, this Cp must be equals to R, right. This Cp must be equals to R. So, calculate Cp. Cp will be three into two that is six minus four into four minus sixteen plus lambda mod divided by under root of A square plus B square that is three square plus four square should be equal to five, okay. And from here we get minus ten plus lambda mod is equals to five into five twenty five. What we need to find? We need to find the value of lambda plus A plus B, okay. So, first calculate lambda from here. So, this will be minus ten plus lambda is equals to twenty five or minus ten plus lambda is equal to minus twenty five. So, from here we get lambda is equal to minus ten. Am I doing some mistake? No. Six minus sixteen minus ten plus lambda, okay. So, lambda is coming out to be twenty five plus ten that is thirty five. And from here we get lambda as twenty five minus ten. Lambda as this will come here or let it move it here. So, lambda is equal to minus twenty five plus ten that is minus fifteen, okay. So, we got lambda as minus fifteen or thirty five. But here it is given that the lambda is positive. So, this value is rejected, okay. So, we got the value as thirty five, lambda is thirty five. So, what will be the equation of the tangent now? It will be three x minus four y plus thirty five is equals to zero. So, this is the equation of tangent because we got the value of lambda and we put in the equation of the straight line. So, now we have to find this point of contact. That means we have to find this P, okay. So, we can do one thing. We can write the equation of normal, right? We can write the equation of normal. So, suppose this is our tangent. So, what will be the equation of normal? Equation of normal means normal is always perpendicular to the tangent, right? So, normal will be perpendicular to this tangent and normal always passes through the center of the circle. So, first let me consider the equation of normal. It will be nothing but four x, this coordinate of x and y will interchange and this sign will change. If it is minus, then it will be positive. That means we have interchanged the coefficient of x and y. So, four x plus three y plus k is equal to zero. This will be the equation of normal, right? And it must satisfy the point c two comma four, right? So, put it in this equation. We get four into two. That is eight plus three into four twelve plus k is equals to zero. From here we got k is equals to minus twenty. From here we got k equal to minus twenty. So, our equation of normal will become four x plus three y minus twenty, right? Is equal to zero. So, we are having the equation of tangent here. Let me tell it as equation one. We are having the equation of normal here. So, this the intersection point of these two, this tangent and normal will give me the coordinates of p, right? So, let's solve this equation one and equation two. So, what can we do? We can multiply the first equation by three. We get three nine x minus twelve y plus one zero five is equals to zero and multiply this equation two by four. We get sixteen x plus twelve y minus eighty is equal to zero. Now, add these two equations. What we get? This will be cancelled out sixteen plus nine. That is twenty five x, twenty five x and plus twenty five, right? Twenty five x plus twenty five equal to zero. So, from here we get x is equal to minus one. x equal to minus one, okay? So, put it in equation one. So, any equation you wish. So, it will be minus three minus four y plus thirty five equal to zero. So, it will become thirty two, thirty five minus thirty two equal to four y or we got y as eight. So, this will be our A and B. That is nothing but A and this is nothing but B. So, we need to find this lambda plus A plus B which is nothing but lambda is thirty five. What is A? A is minus one and what is B? B is eight. So, thirty five minus one, thirty four plus eight that is forty two. So, this will be our required answer, forty two. But in option, we did not get any option as forty two. So, but this will be our answer. No, because what we have done, we have calculated the intersection point of this normal and tangent. So, this must be the P only, the coordinates of P only. So, we got the coordinates of P as minus one and eight. So, this will be our A, this will be our B and what was lambda, lambda was thirty five. Anyhow, we are getting answer as forty two, but no where it is mentioned here. Okay, leave it. Let's see the question number four. It is saying that the tangent which is parallel to the line x minus three y minus two equal to zero of the circle. This, this has point points of contact. Okay. Let me draw a sketch. So, one circle is there, right? One circle is there and tangent which is parallel to the line. So, one equal, line's equation is there, x minus three y minus two. Now, let me draw one tangent parallel to this given line. So, this is our tangent parallel to the given line as points of contact. Okay, we will see now. It is given that the, what is the equation of the circle? It is x square plus, x square plus y square minus four x plus two y minus five equal to zero, right? Minus four x plus two y minus five equal to zero. So, what is the center? Center for this circle is two comma minus one and what is the radius? Radius will be four plus one minus of minus five that is five. So, radius is root ten. Okay. So, this is our center. Two comma minus one and the equation of this line is given as x minus three y minus two equal to zero. Let me remind you, this is not the equation of the tangent. What is given in the question? The equation of the tangent will be something parallel to this line. So, we can write equation of the tangent as x minus three y plus k is equals to zero, right? So, how to find this value of k? Okay, we can do one thing. We can draw a perpendicular from the center. We will draw perpendicular from the center on this tangent and we will equate this as radius. Radius we know of this circle and which is root at ten. So, let me name it as d. Okay. Since we have taken this c, so I am taking this point of contact as d. Okay. So, our c d will be equal to one into two to minus three minus three into minus one will be plus three plus k upon, this will be in mod upon under root of a square plus b square that is one square plus three square. This will be equal to the radius of the circle which is equal to root at ten. So, from here we get mod of five plus k is equal to one plus root ten into root ten that will be ten will be. So, here from here we get, we will get two values of k basically. So, this five plus k is equals to ten. Okay, five plus k equal to ten. From here we get k equal to five and this five plus k is equals to minus ten. From here we get k equal to five plus k equal to minus ten. So, k will be equal to minus ten minus five that is minus fifteen. Okay. So, see basically there will be two lines. There will be two lines which will be parallel to this given line. Since we got the value of k as coming out to be five and minus fifteen. Okay. So, see this will be the scenario like one more tangent will be there. More tangent will be there which will be something like this, right, which will be something like this which will be parallel to the given line. So, whose equation will be like from here we can say let me take it as k is minus fifteen. So, its equation will be three minus sorry x minus three y minus fifteen equals to zero and this line will become x minus three y plus five right plus five equal to zero. So, these two tangents we got, line indicated in the white color. So, what we need to find, we have to find the points of contact, right. So, if suppose this is D, this is E. So, we need to find the point of contacts. How can we do it? Okay, same what we have done in the last question. We will find the value of the, sorry, we will find the equation of the normal like we will find the equation of this CD and we will take the intersection point. We will take the intersection point of normal to these two tangents that will give us the point of contact. Okay. So, we can proceed in that way. So, let me take the equation of normal, sorry, tangent first equation. I am taking this first line, first tangent x minus three y plus five equals to zero. This is the tangent equation of tangent. What will be the equation of normal for this? It will be three x plus y plus k, some k equal to zero, right. This is the one case. And what is the another case? I am taking the equation of tangent as x minus three y minus 15 equal to zero, right. This is one case, the tangent one, and this is the, let me write it as T1. This is as T2. So, now we need to find this value of k and we have to solve both the equations, right. Then we will get the intersection point. This is our approach. So, how can we find the value of k? It's easy. The normal will always pass through the center of the circle. The center of circle is two comma minus one. So, let's put it in equation of the normal. So, we will get six minus one plus k equal to zero. Let me take it as k we have already considered. Let me take it as k one because you may get confused because earlier also we have used the k. So, from here we get k one as six minus one. That is minus five. k one will be minus five. Okay. So, our equation of normal becomes three x plus y minus five equals to zero. So, let me take it as equation one. This as equation two. Now, solve both the equations. So, how can we do? We will multiply this equation two by three and we will add. So, this will become x minus three y plus five equals to zero. Multiply this equation two by three. We will get nine x, right. Three into three. Nine x plus three y minus fifteen is equals to zero. Now add it. So, this will become ten x. Ten x plus five minus fifteen means minus ten. That is equal to zero. So, from here we get x equal to one. Okay. So, if we got x as one, what will be y? The y will be one minus three y plus five equal to zero. So, five plus one six equal to three y. So, we got y as two. So, this is one point of interaction. Sorry, intersection one comma two. So, we were solving with this line. No. So, the coordinates of D will be one comma two. The coordinates of D will be one comma two. Whether one comma two is there in the option? No. We did not have one comma two as option. Okay. So, let's solve these two equations. This tangent to this x minus three y minus fifteen equal to zero. And we know the equation of normal from here. This is three x plus y plus what was given minus five. So, this is minus basically three x plus y minus five equal to zero. So, let's take it as equation three. This is equation four. Now solve it. So, again multiply equation four by three and add both the equations. We will get this multiplied by three. Then nine x plus three y minus fifteen is equals to zero. Now add equation three and four. So, it will get cancelled. Nine x plus x will be ten x. So, we will get minus fifteen minus fifteen minus thirty is equal to zero. From here we get x as three. So, put it in any of the equations. So, x equal to three. This three minus three y minus fifteen is equals to zero. Putting it in equation three. We get minus fifteen plus three. That is nothing but minus twelve is equal to three y. So, y is equal to minus four. So, the coordinates of E will be means coordinates of E will be x three comma minus four. So, two points of contact as they are there one comma two and this three comma minus four. Okay. So, we are having this point in one of the options three comma minus four. So, this option D is correct. Okay. Now, let's take the next question, question number five. It is saying if a circle whose center is minus one comma one touches the straight line. Okay. Then the coordinates of the point of contact are. So, it is given that one circle is there. So, this is our circle. And center is given. Center is known to us for this circle. Touch is the straight line. One straight line is there. So, if this line touches the circle, it means it is the tangent to the circle. Right. And the this perpendicular distance of from this tangent to the center will be equal to the radius of the circle. This is we know. So, we know the center of this circle as let me tell it as a C and coordinates are minus one and one. And the equation of line is x plus two y minus 12 is equal to zero. Right. Then the coordinates of the point of contact are. Okay. So, let me take it as P was coordinates are a comma B. So, we need to find this coordinates of P. Okay. So, how can we do it? How can we do it? We can write the equation of the normal, right? We can write the equation of the normal and we can take the intersection point. So, our tangent is x plus two y minus 12 equal to zero. So, let me assume the equation of normal as two x minus y plus k equal to zero. Okay. Now this normal will pass through the center whose coordinates are more center is minus one comma one. So, let's put the coordinates on this equation of normal. So, we get minus two. Right. Minus two minus one plus k equal to zero. So, from here we get K value of KS. We got the value of KS three. So, our equation of normal will become two x minus y plus three equal to zero. So, this is our equation of tangent which is given in the question. This is what we got the equation of normal. Now solve both these equations we will get the point of intersection that is we will get the point P. Right. So, let's solve and multiply this equation two by two, we will get four x minus two y plus six equal to zero. And this equation is known to us this x plus two y minus 12 equal to zero. So, minus two y plus two y will be cancelled. This will be five x minus six equal to zero. So, we got the value of excess six upon five. So, if we got x put it in any of the equation. So, let me put an equation one only. So, this will be six by five. Plus two y minus 12 equal to zero. Okay. So, what we get we will be having this five six minus 60 equal to minus two y. So, this will be minus 54 by five is equals to minus two y. So, minus minus got cancelled this to 27. So, we got y as 27 upon five. So, this will be the coordinates of point P six comma five and 27 by five. But again we have no options here with these values. But our approach is right now we have we know the equation of tangent, we derived the this equation of normal. Okay, and we solved both the equations. So, this will be the point of intersection. This will be the point of contact. So, the coordinates of P from our end we what we got we got it as six by five comma 27 by five. This will be the answer. So, in this exercise we got this second question I think whose answer is not matching with us. Anyhow, let's see the next question, question number six. This is saying that the area of the triangle formed by tangent at the point a comma B to the circle x square plus y square is equal to R a square and the coordinate axis. So, this is our circle whose center is the origin. Let me draw the coordinate axis also. So, this is our y axis. This is our x axis. Okay. What it is saying the area of the triangle formed by the tangent. Okay, one tangent we have to draw. So, suppose I'm going one tangent here. Right. Okay. First, let me write it. This is our center of the circle. And this is our point P was coordinates are a comma P. So, the area of the triangle formed by the tangent at point a comma B to the circle and the coordinate axis. Okay. Okay. So, let me write it as a the intersection of tangent with x axis as a and the intersection of this tangent with y axis as B. Right. So, we need to find the area of triangle area of triangle. We need to find this area of triangle. Can we find definitely it will be the right angle triangle. Anyhow, if we can find the coordinates of this A and B, then our task is done. So, can we write the equation of tangent at P? We know the equation of circle x square plus y square is equal to R square. So, what will be the equation of tangent at P? It will be xx1 plus yy1 minus R square is equal to 0. Right. Where x1 and yn are the point of contact of the tangent with the circle. So, let me put it in this equation. We get ax plus by equal to R square or minus R square equal to 0. Anyway, any way you can write it now for having the coordinates of a for a we will put y equal to 0. Putting the y equal to 0, we get xs R square upon a. Right. This is the coordinate of x coordinate of this A. This will be R square upon a and the y coordinate on the x axis will be 0 only. And what will be the coordinate of B? For that we need to put x equal to 0 in the equation. So, from there we get y equal to R square upon b. So, the coordinate of b will be 0 comma R square upon b. Right. Now, what will be the area of the triangle ABC? The area of the triangle ABC will be half into base that is R square upon a into R square upon b. That is nothing but R raised to power 4 upon 2 AB. Right. Now, let me remind you. There may be four triangles in this way. We have seen it in the first quadrant. It may occur in the second quadrant also. It may occur in the third quadrant. It may occur in the fourth quadrant. It is better to write it as area as R raised to the power 4 upon 2 into mod of AB. Right. Which is given in our option B. So, this will be the correct answer. Here your triangle ABCs are raised to power 4 upon 2 times mod of AB. Okay. Now see this question number 7. Equation of tangent to the circle which make equal intercepts. Now it is clearly mentioning. Right. It is clearly mentioning that the equal intercepts on the positive coordinate axis. Okay. So, let me draw the sketch for this one. This is our circle. And the equation of the tangent, we need to find the equation of the tangent, which makes equal intercept on the positive coordinate axis. Okay. What will be the center of this circle? Center of this circle will be minus 2 comma minus 2 comma 2. Okay. Let me first, like those gather some information about this circle. So, the equation of circle is x square plus y square plus 4x minus 4y plus 4 equal to 0. So, the center of this circle is minus 2 comma 2. Right. So, this center will be minus 2 comma 2. And what will be radius for this? Radius will be 4 plus 4 and minus 4. So, this will be radius will be 2. Okay. Radius will be 2. Right. Now what we need to find the equation of tangent to the circle which make equal intercept. Okay. So, if I say, I say this is our axis like coordinate axis, this x and y axis, which make equal intercept on the positive side. So, equal intercept means like this doesn't look like equal intercept anyhow. So, this is our tangent to this circle. Right. This is our tangent to this circle. At point, suppose p here, this is the p. This is the p point. Okay. And this equal, this line is making equal intercept on the positive coordinate axis. So, definitely, definitely the slope of this line, the slope of tangent must be slope of tangent. Must be minus 1. Because this will be actually 135 degree. Okay. So, let me assume the equation of tangent as y equal to mx plus c or we can say y equal to m is nothing but minus 1. So, minus x plus c. Okay. So, we have assumed this line as y equal to minus x plus c. Now, how can we find the value of c? We can do one thing. We can drop a perpendicular from the center on this line. Right. That will be equal to the radius. From there, we can find the value of what you say c. Okay. So, it will be, what will be the cp? Like the length of cp will be c upon under root of 1 square plus m square, m is minus 1. So this, and this is equal to the radius. What is radius? Radius is equals to 2. So, from here, we got value of c as 2 root 2. So, put this in the equation of the line. So, our line will be y equal to minus x plus 2 root 2. Right. So, it will, we will get x plus y equal to 2 root 2. So, this is our required equation of the tangent. So, option b is correct. Okay. So, we have the question number 8. It is given if a is greater than 2b, greater than 0 means a and b both are positive. Then the positive value of m for which this line y equal to mx minus b under root 1 plus m square is a common tangent. Okay. To both these circles. So, like equation of two circles are given to us. And question is asking to find the positive value of m for which this line will be a common tangent. Okay. So, let me draw both these two circles. This is our one circle. For circle, the coordinates of center is 0 comma 0 and the coordinates of center for second circle is a comma 0. Okay. And this line is a common tangent to both of the, both the circles. Right. So, if you see, suppose I'm taking this as center one, like circle one. I'm taking this as circle two. So, what will be the coordinates of center for this? It is 0 comma 0. And the coordinates of center for the second circle, c2 is a comma a comma 0. Right. And the equation of this line is given as y equal to mx minus b under root 1 plus m square. Now it is tangent to this first circle also and this is tangent to the second circle also. So, let me draw this perpendicular distance perpendicular from this to this. Okay. So, this will be equal to the radius. This, let me take it as this as a and this as b. This c2b, the length of c2b must be equal to the radius of the second circle. Let me take it as r and this as r. So, this c2b must be equal to r. Okay. And this c2b, how to find this c2b? Okay, we can definitely find out because we know this center as a comma 0 and we know the equation of this length. The equation of the line we can write it as mx. We can write this equation as mx minus y minus b under root of 1 plus m square. So, calculate the perpendicular distance from c2 on this line. So, it will be m into a plus 0 minus b under root of 1 plus m square by under root of m square a square plus b square that is m square plus 1 is equal to radius. Okay. So, what is the radius? The radius is equal to b. Right. So, this will be equal to b. Okay. Now, cross multiply and take the square. So, we will have m a minus b under root of 1 plus m square, whole square is equal to b, b square into 1 plus m square. Right. Now, open it. It will be m square a square plus b square into 1 plus m square minus 2 amd under root 1 plus m square is equal to b square into 1 plus m square. So, since b is positive, we can cancel this term b square a plus m square. So, from here we got this m square a square is equal to 2 amd under root 1 plus m square. Okay. We need to find the positive value of m. Okay. Again, square it. This will be m to the power 4 a to the power 4 that is 4 a square m square p square 1 plus m square. Okay. So, it will be a square a square m square m square. So, we have m square a square is equal to 4 4 4 b square into 1 plus m square. Right. 4 b square into 1 plus m square that is nothing but we are open. This will be m square that a square minus 4 4 b square right 4 b square that is equal to 4 b square. 4 b square plus 4 b square m square. From there I have taken the same square as common. So, from here we get m square is equal to 4 b square upon 4 b square upon a square minus 4 b square. Right. So, from here we get this thing m as 2 b upon under root of a square minus 4 b square. So, this is our required answer. This m square 4 b square upon. Yeah. So, 2 b upon under root of a square minus 4 b square. So, option a is correct. Now let me move to the next question, question number 9. The angle between a pair of tangents from a point p to the circle this is pi by 3 and the locus of p is given as x square plus y square is equal to r square then the value of r is okay. So, we know this is our circle and a pair of tangents is drawn from a point p. Okay. So, let me take it as point p. I'm drawing a pair of tangent. Okay. I'm drawing a pair of tangent on this circle. Right. And we know we need to find the locus of this point p. So, let me do one construction. What I will do, I will do, I will drop a perpendicular, sorry, I will drop a perpendicular on these tangents from the center. Okay. On both the tangents. So, this is what I have dropped and I will join this point p to the center of the circle. Okay. Now let's see what we have done. This is our point p basically. So, this is our point p. Okay. So, let me find the locus of this p point only. So, let me assume the coordinates of ps h comma k. This is our center of the circle which is nothing but 0 comma 0. This is origin. Okay. And this is the pair of tangents. Let me write it as PA and this point of contact as B and what is more given? Okay. The angle between the pair of tangents is given to be pi by three, like this angle. This whole angle is, this whole angle is pi by three. Right. So, if this whole angle is pi by three, this angle APC, APC will be half of this, half of this. That means it will be pi by six. Why? Because the line joining PC actually bisects the angle. It bisects the angle APB. So, this angle will be pi by six. Okay. Now, we need to find the locus of Pino. So, what I will do? I will do one thing. This tan pi by six. What will be tan pi by six? Okay. This is 90 degree, you know. So, tan pi by six will be basically, this will be AC upon PA. Right. Here, if you see, we know the value of AC. Why? Because the AC is nothing but the radius of the circle. What is the radius of the circle? Radius of circle is four. The equation of circle is x square plus y square is equal to four square. So, radius of circle is four. So, AC we know four. Do we know PA, this length of tangent? If you see, yeah, we know the length of tangents also. The length of tangents will be basically under root S1. Where S1 is, where S1 is, what do you say, the power of point, this power of P, power of point P on the circle. Okay. So, S1 will be basically x square, like h square, right, plus k square minus four square. This is S1. Okay. So, the power of point P on the circle is this and the length of tangent, this PA, we take the square root of the power of point as the length of the tangent. And we know the value of this pi by 10 pi by 6. What is 10 pi by 6? It is 1 by root 3. Okay. And what is AC? AC is four. And what is PA? PA is under root h square plus k square minus four square. So, cross multiply it as cross multiply and take the squares on both sides. We will get h square plus k square minus 16 is equal to four square is 16 into three. 16 into three is 48. So, this will become h square plus k square is equal to 64. Now replace this h and k by x and y to find the locus of this point P. So, it will be x square plus y square is equal to 64. And as per question, the locus of P is given as x square plus y square is equal to r square, right. So, what will be r? r will be nothing but equal to eight. Both these represent the same locus of point P. So, r must be equal to eight. So, option D is correct. This is our answer to this question. Option D. Now, let's take the next question, question number 10. It is saying that the normal at the point three comma four on a circle cuts the circle at the point minus one comma minus two. Then the equation of the circle is. Okay. So, this is our circle. Okay. And this is our normal. So, suppose I'm taking this point as A, this point as B. So, A is the coordinates of A is three comma four. We draw a normal at this point on the circle. It cuts the circle at point minus one comma minus two, means the coordinates of B is minus one comma minus two. Then we have to find the equation of circle. Okay. So, normal of a circle always passes through its center. Always passes through its center. Right. So, what will be AB? AB will be the diameter of the given circle. AB will be the diameter of the given circle. And we know the end points. We know the end points of the diameter. So, anyhow, we can write the equation of the circle. This is what we need to do here. So, the coordinates of A is known to us. The coordinates of B is known to us. So, the equation of circle will become, if the x minus x one, x minus x two plus y minus y one into y minus y two is equal to zero. Right. Now put the value of x one and y one and x two and y two in this equation we get x minus three. x minus of minus one that will be x plus one plus y minus y one, that is y minus four into y plus two equal to zero. This is the equation of our circle. So, it will become x square plus x minus three x means minus two x minus three plus y square plus two y minus four y that is minus two y minus eight equal to zero. So, finally, we get x square plus y square minus two x minus two y minus eight equal to zero. This is the required equation of circle x square plus y square minus two x minus two minus seven. So, option B is correct. Now, let's see this next question. The line A x plus B y plus c equal to zero is a normal to the circle. Okay. x square plus y square is equal to r square. This is the circle. The portion of the line A x plus B y plus c intercepted by this circle is of length. So, obviously, this line is a normal. This A x plus B y plus c equal to zero is a normal to the circle. So, the portion of line intercepted by this circle is of length what? This is the circle x square plus y square is equal to r square. Right. This is the equation of normal. It will always pass through the center. And it's both the points of contact will be diameter. So, this will be of length two are like the portion intercepted by this line on the circle will be equal to diameter that will be equal to diameter. So, in last question also we have seen no. This is the circle. And suppose this is our normal. So, what will be this length? This length will be AB will be nothing but the diameter. And what is the length of the diameter? Here the radius is known to be r. Here the radius of this circle is equal to r. So, diameter will be equal to 2r. This will be the length interception, like length intercepted by this circle on this line. So, this 2r will be the answer. Now, let's see this question. If the line A x plus B y plus c equal to zero touches the circle this and is normal to the circle. Okay. So, one line is there, which is the tangent to one circle and that line is normal to another circle. So, we have to find the value of this A and B ordered pair. AB. Okay. So, what we will do? I am taking this circle one circle and I am drawing one tangent to this circle. I am drawing one tangent to this circle, which is normal to one another circle, which is normal to one another circle. So, that circle definitely this line will pass through the center of that circle. So, this is what I have taken. Right. So, this line is given as the equation of this line is A x A x plus B y plus c equal to zero. Okay. And this touches this circle. The equation of this circle is x square plus y square minus two x minus three by five equal to zero. So, the center of this circle is one comma zero. Okay. The center of this circle is one comma zero. And this is the second circle whose equation is x square plus y square plus two x minus four y plus one equal to zero. So, what is the center of this circle? The center of this circle is minus one comma two. Okay. And this will definitely lie on this line. So, let me take it as C one and let me take it as C two. So, the coordinate of C two will be minus one comma two. Okay. So, definitely this line passes through minus one comma two. So, it will satisfy this. It will satisfy this equation minus one comma two. So, put the coordinates of C two on this. So, we get minus A, minus A plus two B plus C equal to zero. This one information we got it minus A plus two B plus C equal to zero. What we need to find? We need to find the value of A comma B. Okay. So, now what I will do now, this perpendicular distance. This suppose this is circle one. So, this will be the equal to the radius one, like equal to C one. Let me name it as P and the C one P must be equal to R one. Okay. So, from here if you see this, what will be C one P? C one P must be equal to radius R one. What will be R one? From here you calculate R one. C one P will be equal to under root of G square one plus F square and minus of C minus of minus 3.5 minus plus 3.5. This will become 5 plus 3 under root 8 by 5. Okay. So, this C one P equal to under root 8 by 5. And now calculate this C one P. It will be A, right. A plus zero plus C, A plus C, mon under root A square plus B square under root A square plus B square is equal to R one. R one is nothing but a root of 8 by 5. Okay. So, from here we get, now we can do one thing. We can replace the value of C here. Let me do it. A, in place of C, I am writing as A minus 2B. A minus 2B, right. So, this mod is equal to under root A square plus B square into under root. 8 by 5. Right. Now I will square on both sides. So, I will get 4. This will be 2A minus 2B whole square. So, 2 into A minus B whole square. So, 2k square 4 then A minus B whole square is equal to A square plus B square into 8 by 5. So, cross multiply it. We will get 5 4s 20, like 5 into 4, 5 into 4 into A square plus B square minus 2AB is equal to 8A square plus 8B square, right. So, it will become 5 into 2. Like this 4 will be cancelled out with this. So, 2, 2. So, we are having 5A square, right. 5A square plus 5B square minus 10AB is equal to 2A square plus 2B square. So, finally we get 3A square minus 10AB plus 3B square is equal to 0. Now, split this minus 10AB as 3 square minus 9AB minus AB plus 3B square is equal to 0. Take 3A common from here. We get A minus 3B minus BA minus 3B is equal to 0. So, A minus 3B either A minus 3B equal to 0 or 3A minus B equal to 0, right. So, either A minus 3B equal to 0 or 3A minus B equal to 0. So, from here we get A equal to 3Bs and from here we get A is equal to or 3A is equal to B, right. So, is there answer available here? So, no option is in the terms of A and B. So, let us take this point option A. It is given that A is 1 if I put this A as 1. So, what will be our B? B will be equal to 1 by 3. But it is 3 here. So, let us check option. Okay, let me put here also. So, if I put A equal to 1 here, we get B equal to 3, okay. So, option A is matching basically. Option A is matching. Similarly, you can check for option B also. This is for option A. Suppose I am checking option B. B it is saying A as 3. If we put A as 3, what will be B? From here if you see A equal to 3, if you put A equal to 3, our B will be A upon 3 means 3 upon 3 that is nothing but 1. So, option B is also satisfying. Okay. And if you put here A equal to 3. So, B will be equal to 3 into 3 that is 9. So, this is not correct. This option is. But this we are getting here. Like A equal to 3B or 3A equal to B. So, hence we are getting A is also as correct as answer and B also as answer. We cannot have this C and D options satisfied by this thing. So, both A and B are correct. But it is better like the answer should be in terms of A and B. Anyhow, this both options are satisfying our what we got in the answer A equal to 3B or 3A equal to. So, this is done here, done with question number. Let's take the question number 13. It is saying so far, so that for all values of theta, this x sin theta minus y cos theta minus 8 touches the circle. Okay. Means here this circle is given. Circle is given x square plus y square is equal to a square and this equation of tangent is given. This equation of tangent is given. What is the equation of circle? Circle is x square plus y square is equal to a square. So, what is radius of this circle? Radius of this circle is equals to A. Okay. And this equation of line is given as this equation of line is given as x sin theta minus y minus y cos theta minus A equal to 0. This is the equation of this tangent. Okay. Now we have to prove that for all values of theta, this will be equal to this line will always touches the circle. So, what we do, we will find the perpendicular distance, the distance from the center to this given line. So, the center is 0 comma 0 for this circle. If we can show that this perpendicular distance, this Cp, suppose I am taking this point of contact as P. If we can say this Cp is independent of theta and this Cp is equals to A, we can say that this circle will always, this, sorry, this equation, this line will always touch the circle. So, let's find the value of Cp. Let's find the length of Cp. So, Cp will be equal to 0 plus 0 minus A comma mod upon under root of A square plus B square that is sin square theta plus cos square theta. So, this is nothing but A upon under root of 1 that is 1. So, Cp is coming out to be A. So, it is not depending on the value of theta. And this is what we need to prove. The length of Cp must be equal to the radius. The length of Cp must be equal to the radius of the circle and that is equal to A. So, yeah, we are getting the value of the length of Cp as A. So, this will always touch the circle independently. And it will not depend on the value of theta. Okay. So, we have proved this. This will be valid for all values of theta. Okay. Now, we have to find the equation of tangent to the circle. This, this, which are parallel and perpendicular to the line. This. Okay. So, like one circle is the one circle is given here. Okay. And one line and one equation of line is given. One equation of line is given. We have to find the equation of tangent, which is perpendicular to this line and which is parallel to this line. Okay. So, let me first identify the center of the circle. So, this is x square plus y square minus 2x minus 4y minus 4 equal to 0. So, center of this circle is 1 comma 2. And what is the radius? Radius will be 1 square plus 2 is 4 minus 4. That will be plus 4. That is radius is root 9 means 3. So, we got the center of this circle as 1 comma 2. The coordinates 1 comma 2 and its radius is 3. Okay. We will see when to utilize this information and the line equation of line is given as 3x minus 3x minus 4y minus 1 equal to 0. Okay. So, now we have to find the equation of tangent, which is parallel to this line. So, equation of tangent, let me write the equation of tangent as which is parallel. First I am taking the case parallel. Okay. I am taking the case one. So, for parallel line to this, I can assume it as 3x minus 4y plus k is equals to 0. Right. And the perpendicular distance from center to this line must be equal to the radius of the circle. So, let's calculate it as Cp. Okay. I am taking this as, let me draw the tangent also. So, okay. This is what I am drawing the tangent to the given circle. Okay. And I am dropping this perpendicular also from the center and I am naming it as Cp. Means point of contact as P. So, this length of Cp must be equal to the radius of the circle. So, what will be the length of Cp? This will be 3 into 1 that is 3 minus 4 minus 4 into 2 that is minus 8. Minus 8 and plus k is come out upon under root of a square plus b square that is 3 square plus 4 square. This must be equal to the radius of the circle, which is nothing but 3. Okay. So, this is what I have assumed this equation of tangent as 3x minus 4y plus k equal to 0. Okay. Parallel to this line. Parallel to this line. So, from here we get minus 5 minus 5 plus k comma r is equal to this will become 5 into 3 15. So, from here we get minus 5 plus k equal to 15 or minus 5 plus k equal to minus 15. So, k is coming to be 15 plus 5 that is 20 and from here k we get k as minus 15 plus 5 that is minus 10. Right. So, our equation of tangent will become two equation two tangents will come similarly one more tangent will be here inside the like parallel to this given line. So, the tangent will become two tangent first I am writing the t1 as 3x minus 4y plus 20 equal to 0 and our second tangent which is parallel to the given line is 3x minus 4y minus 10 equal to 0. Right. So, this is the case of the parallel lines. Now, if you see perpendicular case like now if we see we have to find the equation of tangent perpendicular to this given line. So, I am taking this second case which is perpendicular perpendicular to the line 3x minus 4y minus 1 equal to 0. So, equation of that tangent that tangent will basically that tangent will be perpendicular to this. So, I can assume the equation of that tangent as I will change the coefficient of x and y that is 4x and plus 3y plus k2 is equal to 0. I am taking k here. So, I am taking here k2. So, this will be 4x plus 3y plus k2 is equal to 0. Now, what we have to find this value of k2 similarly we have to calculate the perpendicular distance from center on this line and that perpendicular distance must be equal to the radius of the circle. So, it will be 4. So, it will be 4 into x that is 1 like 4 plus 3 into 2 that is 6 plus k2 mod upon under root of 4 square plus 3 square that is equal to the radius of circle. So, from here we get 10 plus k2 mod is equal to 15. Ok, 10 plus k2 equal to 3 into 5 that is 15. So, from here we get 10 plus k2 is equal to 15 or we get 10 plus k2 is equal to minus 15. So, from here we get k2 is equal to 5 and from here we get k2 is minus 25. So, our equation of tangent will become 4x plus 3y plus 5 equal to 0 and the next equation will become 4x plus 3y minus 25 equal to 0. So, this is the required answer. Now, let's take this question number 15. Find the equation of the family of circle which touch the pair of straight lines. Ok, which touch the pair of straight lines x square minus y square plus 2y minus 1 equal to 0. So, pair of straight lines is given. Let me first solve this equation. The pair of straight line is given as x square minus y square plus 2y minus 1 equal to 0. This is nothing but x square minus y square minus 2y plus 1 means y minus 1 whole square is equal to 0. So, we got the two line as x plus y minus 1 equal to 0. This is the first line and we got x minus y plus 1 equal to 0. This is the second line. Now, let me plot these two lines. So, suppose this is our y axis, this is our x axis. What is the point of intersection for these two points? If we put x equal to 0, y1, 0, y1. So, this 0 comma 1 is the point of intersection of these two lines. I am going to draw these two lines. So, these two lines are given here. This will be with positive slope. This line will be with positive slope. That means this line is x minus y plus 1 equal to 0. And this line is basically x plus y minus 1 equal to 0. And both these points meet at 0 comma 1. Like this point is basically 0 comma 1. This point is 0 comma 1. 0 comma 1. Let me write it. This point is 0 comma 1. Now, we have to find the equation of family of circles which touch the pair of straight lines. Which touch the pair of straight lines. Now, let's observe, we can have these type of circles. Like this may be one circle. Now, see this circle is touching both these pair of circles, both these pair of lines. Okay. And circle can be here also. So, we have to find these family of circles which are touching both these straight lines. So, look, circles can be in this way. But one thing you must observe, like circles, the center of the circles, this, suppose I am taking it as a, okay, let me, let me draw this line also. So, the point of intersection of this is 0 comma 1. Okay. So, yeah, I need to observe you, like observe these things. Like the center of these circles will always be on the y axis. Of all the family of circles, it is not like that. The circle can be of bigger size also. And there will be infinitely many circles in this way. So, but one thing will be common. The circle for these points will be on the center of these circles, this one, suppose I am taking this one and three. The center of these circles will be x coordinate will be 0. Right. And y coordinate will be root two times of this square root of, sorry, what do you say? I think let me explain you within this method in this way. Like, what is, okay, let me ask you what is the angle of intersection like what is the angle between these two lines. This will be 90 degree, basically. This will be 90 degree. Okay. So, let me do one thing. What I am doing? I am joining these two points. Okay. I am joining these two points. So, this will be r, right. This will also be r. And this will be r, this will be r. And this distance if you see, from this point, 0, 1, if you see, this all the centers of these circles will lie on 0, r by root two. Like, 0, r by root, r by root two. No, r into root two. R into root two. Similarly, if you observe centers of these, it will lie on this r root two comma the y coordinate of these circles will be 1. Right. So, y coordinates will be r root two comma one. Hope it is clear to all. Like, there will be a group of circles in this way. In which the x axis, like the x coordinate of the circles, which are on this one and three side will be 0 and the y coordinate for these circles, which let me mark it as 2 comma 4. The y coordinate of the center of this will be fixed that is one and the x coordinate will be a function of the distance from this that will be r root two. Right. So, hope this is clear to all. Now, let's take this next question, question number 16. It is saying that they find the value of lambda so that the line may touch the circle. So, this equation is there. This equation of circle is x square plus y square minus 4x minus 8 y minus 5 equal to zero and one equation of line is there. That is 3x minus 4 y minus lambda is equal to zero. So, the question is asking to make this line as a tangent to the circle. So, what is the center for this center for this circle is 2 comma 4 and what is the radius radius is 4 plus 16 minus of minus 5 that is 15 so radius is coming out to be 5. So, distance from center to this line must be equal to the radius. So, this will be 3 into 2 minus 4 into 4 minus lambda mod upon under root of a square plus b square that is 3 square plus 4 square this must be equal to the radius. So, this will be 6 minus 16 minus lambda mod is equal to 25. So, from here we get minus 10 minus lambda mod equal to 25. So, minus 10 minus lambda equal to 25 or minus 10 minus lambda equal to minus 25. So, from here we get lambda equal to 10 minus 25 that is minus 35. So, this is one value of lambda what we got and from here we get 25 minus 10 that is 15 is equals to 25 minus 10 is 15. So, we got the two values of lambda here minus 35 and 15. So, this is the value of lambda. I think we are done with this chapter. Ok, one more problem is there. It is saying that the so that the area of the triangle formed by positive x axis the normal and the tangent to the circle at point 1 comma root 3 is 2 root 3. Ok. So, the question is saying one this is the circle whose center is the original itself. So, this is our coordinate axis y and x. Ok, one point is given here 1 comma root 2 and we are drawing one normal to this point and we are drawing one tangent to this point. Suppose this is our tangent and this is our normal. Normal will always pass through the center. So, this is what is given. So, suppose I am taking this point as p whose coordinates are 1 comma root 3. Ok. And this is our normal. This is our tangent at that point and this is our origin as the center of the circle. So, what it is asking area of the triangle formed by positive x axis positive x axis normal and tangent. So, we need to find the area of triangle. Area of triangle p o. Ok. Let me name it as a. Right. We need to find the area of triangle p o a. p o a. Ok. So, this if you see this o p is equal this will be no doubt this will be 90 degree. So, this will be a right angle triangle. So, our area will be basically half into o p into p a. Right. So, we know the value of this o p. o p is the radius of the circle. What is the radius of circle here? The circle is x square plus y square is equal to 4. So, radius is 2 here. So, we do know the value of o p that is 2. Now, we need to find this value of p a. So, p a if you see p a is nothing but the length of tangent. Right. Length of tangent which we can find it out by or if we see this if we consider this point o sorry this point a as point outside the circle. Right. As point outside the circle we can definitely find the value of p a as the length of tangent which is nothing but let me let me first write the equation of tangent at p that will be I think helpful. Let me write the equation of tangent at point p. So, our circle is x square plus y square minus 4 equal to 0 and I am writing the equation of tangent at p that is x x 1 plus y y 1 minus 4 equal to 0. Now, I will put this 1 in place of x 1 and root 3 in place of y 1. So, it will be x plus root 3 into y minus 4 equal to 0. So, this is our equation of tangent basically. This is equation of tangent. Okay. Now, to find the coordinates of a to find the coordinates of a what I will do I will put a y equal to 0. Right. For finding the coordinates of a I will put y equal to 0 from here we get x equal to 4. So, coordinates of 4 will be basically coordinates of 4 will be 4 comma 0. Right. So, this will be the coordinates of a and now we can find the length of p a. We can find the length of p a. Right. So, this approach was not helpful I think this was the better approach like we have find the equation of tangent then we find the coordinates of a. Now, the value of p a will be, means length of p a will be 4 minus 1 whole square that is 9 plus root 3 whole square that is 3 that is nothing but root 12 or we can say 4 into 3 that is 2 root 3. So, p a is 2 root 3. So, this 2 2 will be cancelled out our area is coming out to be 2 root 3, which is what we have to prove. So, hence it is proved. So, this is all I think for this exercise. Yeah, this is the last question. So, this video. Like, I think this is of bigger length but can't help this exercise was big and this was like the questions solving take two time basically. So, anyhow, we are done with this exercise number 4. We will see you again with the next exercise that is exercise number 5 of the circles. Till then, Tata, goodbye, take care.