 Hello, everyone. So, we have now entered in this coordinate geometry section. Basically, you can say. So, in last class, we discussed about circles, some basic questions of exercise one. So, today we will move forward to the next exercise exercise two of the circles. So, let's start. So, this is our first question. It is saying if the line x plus two lambda y plus seven equal to zero is a diameter of the circle, x square plus y square minus six x plus two y equal to zero, then the value of lambda is. So, one circle is given, and it's diameter is also given the equation of the diameter basically. So, we have to find this value of lambda. Okay. So, let me write the equation of a circle. It is x square plus y square minus six x plus two y equal to zero, right? So, we can identify the center of the circle, right? Center will be minus g comma minus f. So, we can say three comma minus one, right? Three comma minus one because two g is minus six here. So, g becomes minus three and two f is two here. So, f becomes one. So, our center is minus g means minus of three is three and minus of f means minus one. Now, this is the center of the circle, right? Now, it must lie on the diameter and we know the equation of diameter. Diameter is given as x plus x plus two lambda y plus seven equals to zero. Now, this point must satisfy this. So, put the value of x and y here. So, three plus two lambda, what is y minus one plus seven is equal to zero. So, this will become ten minus two lambda is equal to zero. From here, we get lambda is equal to five. Okay. So, this is our answer. Lambda equal to five. So, option C is correct. Let's see the next question, question number two. If one end of diameter of the circle, this is minus one comma two, then the other end of the diameter is. Okay. So, equation of circle is given. It's one end of diameter is given and we are asked to find its, find the other end of the diameter. We have to find the coordinates of other end of diameter. Okay. So, let me write first the equation of circle. It is saying two x square plus two y square minus four x minus eight y plus two is equals to zero. So, okay, let me first make the coefficient of x square and y square as one. I have to divide by two. So, it will be x square plus y square minus two x minus four y plus one is equals to zero. Okay. Now, we can say the center of this circle. What will be the center? That will be one comma two. Right. One comma two will be the center of this circle. Now, let me draw the circle itself. Let me draw its diameter also. Now, one end, let this AB be the diameter. Let this AB be the diameter. We know the one end of diameter that is minus one comma two. Okay. And we have to find the other, the coordinates of this B. So, let me assume it as A comma B. And we know the center of this circle. What is that? One comma two. The center of this circle is one comma two. So, if AB is the diameter and this is center C. So, this AC will be equal to BC. Right. So, C will be the midpoint of A and B. So, we can say this AC must be equal to BC. Okay. Both are radius only. So, AC equal to BC. Now, I will, how can I find the midpoint of this AB? This will be minus one plus A upon two. And that is nothing but that is equal to one. Okay. And this two plus B upon two is equal to two. Right. The coordinates of this C. So, from here I get minus A plus one is equals to two. Or we can say A is equal to three. And what will be B from here? Two plus B is equals to four. So, B will be equal to plus four minus two, that will be two. So, three comma two. Three comma two will be the coordinates of B. So, this will be our correct answer. Option number B. Right. So, moving ahead, let's see question number three. It is saying if a circle passes through the points, zero comma zero, A comma zero and zero comma B, then the centers, then the center of the circle is. Okay. Means the circle is passing through origin basically. So, let me draw this sketch, what is given in the question. This is our circle. And let me consider this as coordinate axis. Okay. So, this will be the zero comma zero point. Sorry. This will be zero comma zero point. This point will be A comma zero. And this point will be zero comma B. Right. And what it is saying, we have to find the center of the circle. Okay. We have to find the center of the circle. So, center of the circle will be basically, if you say this will be our diameter. Sorry. So, this point, this, suppose I am taking it as B and this is as A. So, AB will be our diameter. Right. And what will be the midpoint of this? The midpoint of AB will be the center of the circle. So, we can easily find the coordinates of this C. C will be nothing but A plus zero by two. And the coordinates, the X axis, sorry, X coordinate of C will be A plus zero upon two. And Y coordinate will be zero plus B upon two. So, X, we can give the coordinates of C as A upon two comma B upon two. Okay. Why this is, why this AB will be diameter because it is making this AB is sustaining the 90 degree at the circumference. Right. Why 90 degree? Because both these are, what you said, both this, this is Y axis, our Y axis, we can consider this as Y axis and this as X axis. Hence, this AB will be the diameter and C will be the midpoint of this AB. And by this, we find the coordinates of C as A by two comma B by two. Okay. So, let's take the next question, question number four. A circle passes through the points minus one comma three and five comma eleven and its radius is five. Then its center is. Okay. So, it is saying the question has provided two points from which the circle is passing and we also know the radius of that circle. Okay. So, suppose this is our point A whose coordinates are minus one comma three and one more point is given as B whose coordinates are given as five comma eleven. Okay. Circle is passing through these two points and we also know the radius of that circle is as five. These three things are given in the question and now we have to find the center. Okay. One, like the radius is known to us, radius is given as five. So, let me check the distance of AB, like what is the length of AB? Okay. Like one intuition is coming in my mind like it may come as 10. Why? It may, and if it comes as 10, then this AB will be the diameter. Right. So, once we can arrive at that thing, then we can say the midpoint of this AB will be the center of this circle. So, let me first find the distance, this length of AB. The length of AB will be under root five minus minus one pole squared plus eleven minus three pole squared. That is six squared thirty six plus eight squared sixty four under root. So, yeah, it's true. The AB is coming out to be ten unit. AB is ten unit. It means AB is the diameter. Right. AB is the diameter and if AB is the diameter, AB is the diameter of the circle. AB is the diameter. So, center will be midpoint. What will be the center midpoint of this AB? That means minus one plus five upon two comma three plus eleven upon two. This will be the coordinates of the center. That will be five minus one four by two. That will be two comma seven. Eleven plus three, fourteen by two, center. So, this will be the coordinates of the center. Two comma seven. So, option C is correct. So, we are done with this question. Question number four is done. Now, let's move to question number five. Now, it is saying the radius of the circle having center at two comma one, whose one of the chord is a diameter of the circle, this. Okay. So, let me draw a sketch for this question. It's saying one circle is there. So, this is our circle. The radius of the circle having center at two comma one. Okay. We will draw it for the, whose one of the chord is the diameter of the circle. Okay. So, I'm drawing, going to draw one more, one more circle here. This is the chord of the bigger circle, right? And I think this much sketch is sufficient. We can move with our question now. So, it is saying the radius of the circle having center at two comma one. So, I'm taking this center as O suppose and this coordinates of this O is two comma one, whose one of the chord, like this AB is the chord of the, this circle. So, this AB is now behaving as the diameter of the circle. Means, it is diameter, this AB is the diameter of this circle. The smaller circle and its equation is given by X square plus Y square minus two X minus six Y plus six equals to zero. Okay. So, what we have to find? We have to find the radius of this bigger circle. We have to find the radius of the circle whose center is two comma one, whose one chord is the diameter of the circle. This. Okay. So, this AB is diameter of this bigger, what you said, this given circle's equation means a smaller circle. So, what is the center of this? What is the center of this circle? The center of this circle will be one comma three. Center of this circle will be one comma three and it will lie on AB itself. Okay. Suppose I'm taking this as point as M. So, the coordinates of this will be one comma three and what is the radius for this circle? Radius for this circle will be one is squared plus three is where minus C minus C is minus six. So, one plus nine ten minus six equal to root four. So, radius of this is two. So, this MB we can say, this MB or what you say, AM is equals to two. Okay. So, this unit we came to know at two. And what we need to find? We need to find the radius. Okay. So, let me draw that also. This we need to find it. We need to find this OB. This OB will be the radius of the bigger circle. So, we need to find OB and we know this distance as MB is two. So, we can apply Pythagorean theorem here and we also can calculate this distance of OM. So, what will be OM basically? What will be the length of OM? It will be two minus one is square because we know the coordinates of O and M both. So, two minus one whole square plus one minus three whole square under root. So, this will be one square, one plus minus two square that will be four. So, this will be under root five. Under root five. So, we know OM and we know MB. So, we can entriangle OMR, entriangle sorry, entriangle OMB. We can apply Pythagorean theorem. So, this will be OM square plus MB square is equal to OB squared. And this OB is what we need actually. So, OM is OM squared is five plus MB squared. What is MB squared? MB squared is four that will be equal to OB squared. So, nine is equal to OB squared or we can say OB is equal to three. So, OB is nothing but the radius of the bigger circle that will be three. Three unit, right? So, our answer is option A, three unit. So, let's see the next question. The center of the circle inscribed in the square on y the lines, x is square minus 8x plus 12 equal to zero and y is square minus 14y plus 45 equal to zero. Okay. So, we will draw the sketch for this. Later, this will be circle and this be our circle. The circle is inscribed. So, it will be live within the square. This will be basically our figure. The center of the circle inscribed in a square formed by lines. We have to find the center of the circle. Now, these squares are formed by the line x square minus 8x, x square minus 8x plus 12 is equals to zero. And y square minus 14y plus 45 is equals to zero. This consists of two lines basically. So, x square, we can factorize it as x square minus 6x minus 2x plus 12 is equal to zero. From here, we get x, x minus 6 minus 2, x minus 6 is equals to zero. So, x is equal to 2 and x is equal to 6. These two solutions we are getting from here. And from here, we will get y square minus 5, 9s are 45. So, 9y minus 5y plus 45 is equals to zero. So, yy minus 9 minus 5y minus 9 is equals to zero. From here, we get y equal to 5 and 9. So, basically, if you see what, let me name this square as a, b, c and d. So, this a, d line will be basically x equal to 2, right? And this c, d line will be, this c, d line will be actually what you say, x is equal to 6. Okay, this x equal to 2 and this is x equal to 6. And what will be this a, b line? This a, b line will be basically y equal to 5 and this dc line will be y equal to 9. So, our, what will be the length of the sides of the square? It will be 4. This will also 4. This will be also 4. And this will also be 4. Okay. So, now we have to find the center of the circle. So, if I tell what will be the coordinates of a, b, c, d. So, we can say this will be 2, 5. The b coordinates of b will be 6, 5. Coordinates of c will be 6, 9. And coordinates of d will be 2, 9. Okay. And now we have to find the center of the circle. So, it will be nothing but it will be half of this. Means x will be 2 plus 6 upon 2, like if we say c center. So, center will be this 2 plus 6 upon 2 and this 5 plus 9 upon 2. 5 plus 9 upon 2. That will be 4 comma 8 by 2, 4 comma 7. 9, 5, 14 by 2. So, 4 by 7 will be ours. So, coordinates of 4 will be, sorry, coordinates of center will be 4 comma 7. So, option a is correct. Now, coming to the next question, a, b, c, d, a square plus side is a. The equation of circle, circumscribing the square, taking a, b and a, d as the excess of reference. Okay. In last question, the circle was inscribed. Now, in this question, it is saying that the circle is circumscribing the square. So, let me draw a sketch here also. So, this is our square and let me draw a circle, which is circumscribing this. Let me draw it separately. First, I am drawing this circle and I will draw a square inside it. Right? So, what is given in the question? This a, b, c, d, j, a, b, c, d is a square. a, b, c, d is a square whose side is a. Okay. The equation of circle, we have to find the equation of the circle, which is circumscribing the square and taking a, b and a, d as the excess of reference. Okay. So, we have taken a, b and a, d as the excess of reference. So, suppose I am taking the coordinates, taking the a as origin. So, the coordinates of a will be 0, 0. And what will be the coordinates of b? Coordinates of b will be a, 0. Why? Because the sides of the square is of length a. And what will be the coordinates of this d? Coordinates of d will be 0, a. And the coordinates of c will be a, a. So, now we have to find the equation of the circle. So, how can we find? How can we find? Let me join this line segment bd. Let me join this line segment bd. So, what will be the bd? Bd will be the diameter. Bd will be diameter of this circle. Bd will be diameter of this circle. And we know the coordinates of b and d. So, we can write the equation of circles in a diametric form as x minus x1, x minus x2, plus y minus y1, y minus y2 is equal to 0. Okay? If we know the coordinates of the ends of diameter, we can write the equation of circle in this form. So, this is called diametric form. We can say it as diametric form, diametric form of circle. So, equation of circle is written in this way if we know the ends of the diameter. So, what is x1 and x2? So, let me tell this b as x1 and y1. And let me tell this b as x2 comma y2. Okay? So, our equation will become x minus a and x minus x2. What is x2? x2 is 0. So, it will be x only. Then y and minus y1, y1 is 0 here. So, it will be y into y minus a equal to 0. This will be the equation of circles. So, it will become x square minus ax plus y square minus ay is equals to 0. So, this is our required equation of circle. x square plus y square minus ax minus ax minus ay equal to 0. So, option c is correct. Option c is correct answer about this. Let us take the next question, question number 8. It is saying the locus of the center of the circle for which one end of diameter is 3 comma 3 while the other end lies on the line x plus y equal to 4. While the other end lies on the x plus y equal to 4. Okay? And we have to find the locus of the center of the circle. So, we are given with one circle. We are given with one circle. This is our circle. And one end of diameter is also given. So, let me draw diameter of this. So, suppose this is our diameter of this circle. And the one end is one end of diameter is known to us 3 comma 3 and the other end lies on the line. Okay? So, let me draw one straight line here. Yeah, I think sketch is done. We can now this AB, let me assume this as AB. This AB is the diameter. This will be the center. The coordinates of A is given as 3 comma 3. And this B, the other end of diameter lies on the line. The equation of line is given as x plus y equal to 4. Okay? So, B lies on this line. So, what can I take the coordinates of B? Suppose I am taking coordinates of B as A, the x coordinate as A. Then what will be the y coordinate? Y coordinate will be 4 minus A, right? 4 minus A. Since it lies on the line, x plus y equal to 4. Now, what we need to know? We need to find the locus of the center of the circle. So, let me assume this center as C. And let me consider its coordinates as h comma k. Okay? So, A coordinates of A is known to us. Coordinates of B is known to us. Means, it is in terms of A, but it is known. So, we can say, find the coordinates of h, means coordinates of C. So, what will be the coordinates of C, basically? Coordinates of C is h comma k we have considered. So, h will be actually 3 plus, it will be a midpoint of A and B. So, this will be 3 plus A upon 2, right? And what will be our k? K will be actually midpoint of, sorry, 3 plus 4 minus A. 3 plus 4 minus A. Sum of both the coordinates divided by 2. So, 3 plus 4 minus A upon 2. That is equal to 7 minus A upon 2. That comes out to be the value of k. So, what we will do now? We will write A in terms of h and k. And we will then equate what we got, the value of A. So that, and finally we will replace the h and k by x and y. So, from here we get 2h minus 3 is equals to A. 2h minus 3 is equals to A, right? And from here we get 2k, 2k minus 7 is equals to minus A. Or we can say A is equals to 7 minus 2k. Okay. So, from here we can say 2h minus 3 is equals to 7 minus 2k. Why? Because 2h minus 3 is also equals to A. And 7 minus 2k is also equals to A. So, from here we get 2h, 2h plus 2k is equals to 10. Okay. 2h plus 2k, 7 plus 3, 10. From here we get h plus k is equals to 10. Now replace h and k by x and y. So, x plus y is equals to, oh, this will be 10 upon 2. So, 5. So, now replace h and k by x and y. So, we will get x plus y equals to 5. So, the locus of the center of the circle will be a straight line basically. x plus y equals to 5. So, our answer will be option B. Now, let's see the next question. Question number 9. The equation of the circle which passes through 1 comma 0 and 0 comma 1 and has its radius as small as possible. So, two points are given from where the circle is passing and we have to say its radius as small as possible. So, we have to draw one circle. Let us understand this by, let me draw it first. Let me mark it then I will explain. Let me mark this as A, this as B and this coordinates of this as 1 comma 0 and coordinates of B as 0 comma 1. Right. So, this coordinates of A and B are given means of any two points is given. And the question is saying that they find the equation of the circle having its radius as small as possible. So, this yellow colored circle is the smallest circle passing through A and B having A, B as the diameter. Means what I am saying, if two points are given and we have to find the least radius circle, radius of least, sorry, circle of least radius, then A, B that those two points must be the diameter of that circle. Why? Because now let's see this white circle what I have drawn here. Let's see this white colored circle. So, for this is white colored circle is also passing through A and B. Okay, no, but it is not the smallest circle passing through A and B. There will be actually infinite number of circles, circles passing through A and B, but the smallest circle will be that which will be having A, B as the diameter. So, this is the thing in this question. This is what you have to understand. So, I have to write the equation of this circle basically. And we know the diametric ends of this circle. So, we can write this as x minus x1 into x minus x2 plus y minus y1 into y minus y2 is equals to 0, where x1, y1 and x2, y2 are the diametric ends. So, let me write it as x minus 1 into x plus y into y minus 1 is equals to 0. So, this will be x square, x square minus x plus y square minus y is equals to 0, or we can rewrite it as x square plus y square minus x minus y equals to 0. So, this is our equation of the circles. x square plus y square minus x minus y equals to 0. So, option B is correct. So, we are done with this question number 9. Let me take question number 10. Now, what do you say? If the points 1, 2, 3, 4, like 4 points are given with their coordinates and it is saying that the 4 points are consyclic, then the value of c. Consyclic means all these 4 points lie on a circle. All these 4 points lie on a circle. So, let me draw and this point, we consider it as A, B, C and D. So, what are these points? 2 comma 0, 2 comma 0. This will, this is 0 comma 1. This is 4 comma 5, 4 comma 5. And this is 0 comma c. This is 0 comma c. So, let me assume this as a center O. So, if you see the distance of all these points from this center will be equal basically and all these will be equal to R. So, all these distances will be equal. That is OA is equals to OB is equal to OC is equals to OD and that will be equal to radius, equal to the radius of the circle. Now, let me find this OA first. So, what will be OA? Or we can say square it. So, OA square is equal to OB square is equal to OC square is equal to OD square. We can say this also. So, we will get rid of this under root thing. So, what will be OA squared? OA squared will be OA squared. Okay. Then in that case, I have to take the, take this as, I have to assume some coordinates of O, no. So, let me take it as a comma B. We can do no. Yeah, we can do because C is here. So, we can take it as a comma in alternate issue. So, OA squared will be A minus 2 square plus B squared. Right? And that will be equal to OB squared. What will be OB squared? That will be A squared plus B minus 1 square. Right? So, from here we get A squared plus 4 minus, A squared plus 4 minus 4A plus B squared is equal to A squared plus B squared plus 1 minus 2B. So, A squared B squared will be cancelled out. A squared B squared will be cancelled out. This will be 4 minus 4A minus 1 plus 2B equal to 0. Or we can say 2B minus 4A 2B minus 4A plus 3 equals to 0. So, this is our first vacation. Let's consider it as first equation. Now similarly, we can say this OB squared is equal to OC squared or OA squared is equal to, OA squared is equal to OC squared. So, what is OA squared? OA squared is A minus 2A squared plus B squared is equal to OC squared. And what will be OC squared? A minus 4 squared plus B minus 5 whole square. So, A squared plus 4 minus 4A plus B squared is equal to A squared plus 16 minus 8A plus B squared plus 25 minus 10B. So, A squared B squared is called cancelled. Here we are left with 4 minus 4A and plus 8A plus 10B and plus 16 and plus 25, 35, 41, minus 41 is equal to 0. So, further on further simplification, it becomes 8A minus 4A means 4A plus 10B, 4A plus 10B minus 41 plus 4, minus 41 plus 4 becomes minus 37, right? A is equals to 0. So, this is our equation number 2. From here, we can find the value of A and B, okay? And after finding the value of A and B, what we will do? We will again equate this OD equal to OA or OD equal to OB in A and we will find the equation in C. From there, we can find the value of C. So, this is the motive behind doing all these exercises. So, let me solve it. So, if I add both these equations, this 4A term will be cancelled. So, it will be 2B plus 10B. That will be 12B, right? 12B minus 34 is equals to 0. So, B will be 34 by 12. So, that will be 17 by 6, right? B will be equal to 17 by 6. So, what will be our 4B? 4B will be 2 into 17 by 6 plus 3 is equals to 4A, right? So, this will be 17 by 3 plus 3 is equals to 4A. So, 4A will be equal to 317 plus 320. These three 3s are 9s. Means 17 plus 9 will be 26, right? So, 17 plus 9. 4A will be 317 plus 9, 26, 26 by 3, 26 by 3 or we will get A as 26 upon 12 or 13 upon 6. So, A comes out as 13 upon 6. So, this will be the coordinates of the center of the circle. This O comes out to be 13 upon 6 and 17 upon 6. Now, again equating this value, this OA is equals to OD. We can find the value of C. We can find the value of C from here, okay? You can also equate this OB is equals to OD not an issue. Then also you can find the value of C. So, I think it is clear. I am not taking this forward. So, I am moving to the next question now. So, that will be equal to question 11, okay? So, it is saying the point on a circle nearest to the point 2 comma 1, okay? One point is given. And it is saying the point on the circle nearest to that point is at distance 4 unit. And farthest point is 6 comma 5. Then the center of the circle is, okay? So, if you see, now it is saying this is, suppose this is the point P. This is the point P 2 comma 1, okay? So, let me consider this as point A. Means on extending this line, it meets the circle at A and B. And A is the nearest point. It is at distance 4 units from here. It is at distance 4 units from here. And the farthest point is 6 comma 5, okay? And here one concept is there, like the farthest distance from P will be along the diameter, actually. If PA is the sort, if PA is 4 unit, okay? And A is the point nearest to the P. So, the farthest point will be along the diameter itself. What does it mean? It means that AB must be the diameter. Anyway, we can say the P point will be farthest from P, right? So, PA is given out to be 4. This PA is given out to be 4. And the coordinates of point B is given as 6 comma 5, okay? Then the center of the circle, we need to find the center of the circle, okay? So, this distance is given as 4. So, PA is given as 4. And farthest distance is 6 comma 5. We can find the distance, this PB. We can find this distance PB. What will be PB basically? PB will be 4 square, 6 minus 2 square plus 5 minus 1 square, 4 square under it. So, 16 plus 16, that will be 4 root 2. This PB will be 4 root 2. This PB will be 4 root 2. So, we can say AB will be 4 root 2 minus 4, right? That is nothing but taking 4 comma root 2 minus 1 is AB. That is the length of AB will be 4 into root 2 minus 1, okay? Now, if you see this total distance, this total length PB, right? And A point, if you see, we can say this A point is dividing the line segment PB in the ratio of, from here you see one thing. PA is 4, no? This PA is 4. This AB is, let's understand in this way. This PA is 4. And this, what do you say? AB is 4 into root 2 minus 1, right? So, this PA upon AB. How much it will be? What is PA? PA will be length of PA will be 4. Length of AB will be 4 into root 2 minus 1. So, it will be 1 upon root 2 minus 1. This 4 and 4 will be cancelled out. So, actually this point A is dividing this line segment PB in the ratio of this 1 is to, 1 is to root 2 minus 1. Is it clear? This point A is dividing this line segment PB in the ratio of 1 is to root 2 minus 1. So, we can find the coordinates of A. What will be the coordinates of A basically? This 1 into 6, right? In the section formula we have learnt, no? mx2 plus nx1 plus root 2 minus 1 into this x upon m plus n means 1 plus root 2 minus 1. And what will be the y coordinate of A? It will be 1 into 5, which we can write 5 plus root 2 minus 1 into this thing 1 upon 1 plus root 2 minus 1. This will be the coordinates of A. So, 6 plus 2 root 2 minus 2 upon 1 minus 1 cancelled out root 2 comma 5 minus 1 4 plus root 2 upon 1 and 1 cancelled out upon root 2. So, finally we can write the coordinates of A as 4 plus 2 root 2 upon root 2 and 4 plus root 2 upon root 2. Now, we know the coordinates of A. We know the coordinates of B. Then what will be the coordinate of C? It will be the midpoint, right? It will be the midpoint of this. So, if we say C is the midpoint, C is the midpoint of AB. If C is the midpoint of AB, so it must be equal to the coordinates of C must be. So, let me write it as coordinates of C as A comma B. So, A will be equals to x coordinate of this 4 plus 2 root 2 upon root 2 plus x coordinate of B that is 6 whole divided by 2, right? This will be A. And what will be B? B will be y coordinate of this A. That means 4 plus root 2 upon root 2 plus y coordinate of this, what do you say? Our point B, what was that? It's 5, right? So, that will be plus 5 upon. So, this will be our coordinates of the center of the circle, okay? So, on simplification, I guess we get this option as C, 4 plus root 2 comma 3 plus root 2. So, this is our option C. The answer to this question is C, I think. Take the next question, question number 12. The intercepts on the line y equal to x by the circle, x square plus y square minus 2x equal to 0 is AB. Equation of the circle on AB as a diameter is, okay? Let me first draw this sketch. It will be more convenient. Line what it is saying, line y equal to x, no? Suppose we are taking this line y equal to x, okay? This line is y equal to x and this is the circle. So, the intercept on the line y equal to x by the circle. This x square plus y square minus 2x. The equation of this circle is x square plus y square minus 2x equals to 0 is AB, right? This is AB actually, okay? And the equation of circle is given as x square plus y square minus 2x equal to 0. Now, we have to draw one circle, equation of circle on AB as the diameter. So, we have to find the equation of that circle whose diameter is AB. Basically, this thing, this is what is asked in that question. Now, taking AB as diameter, let me draw one circle. Yeah. So, we have to find the equation of this circle, basically. We need to find the equation of this circle. Whose AB is the diameter? AB is the diameter. AB is the diameter for this circle. Okay. So, in the equation of this, let me put y equal to x because y equal to x also lies on the circle. So, if you put y equal to x, we will get x square plus x square minus 2x is equals to 0 or 2x square minus 2x is equals to 0. And again, putting x equal to y, again, here in this equation, we get y square plus y square minus 2y is equals to 0. Actually, what I am trying to do, I am trying to identify the coordinates of A and B. Like, I am solving these two equations now. So, we will get the point of intersection. So, this is what? This 2y square minus 2y is equals to 0. So, solving this quadratic in x, 2x square minus 2x equal to 0 will give me the x coordinate of A and B. From here, we get x1 and x2. That will be the x coordinate of A and B. And from here, we will get y1 and y2, which will be the y coordinate of A and B. Okay. Then, after getting that x1, x2 and y1, y2, we used to write it in form as x minus x1 into x minus x2 plus y minus y1 into y minus y2. So, instead of doing all these things, if we add these two quadratic now, this 2x square minus 2x plus 2y square minus 2y is equals to 0. This will be the required equation of the circle. This will be the required equation of the circle. Why? I am directly adding it. I am directly adding it. Why? From these two equations, we get two values of x. Okay. And that two values will be the x coordinate of A and B. And from here, we get two values of y, y1 and y2. That will be the y coordinate of A and B. Okay. Then, after finding this x1 and x2, I will again make a quadratic by putting x minus x1 into x minus x2 plus y minus y1 into y minus y2. So, instead of doing all these things, I am just adding these two quadratic. Adding these two quadratic, we get the required equation of the circle. So, you can take two common from here. So, x square plus y square minus x minus y equal to 0. This will be our answer. x square plus y square minus x minus y equal to 0. So, this question number 12 is this option A is correct for. Question number 12. So, what is given? Question number 13 is saying, find the equation of circle. The int points of whose diameter are 2 comma minus 3 and minus 2 comma 4. So, it's easy. We have done similar type of questions earlier also. So, 2 comma minus 3 and B minus 2 comma 4. These are the diametric ends. Find the equation of circle. The ints of whose diameter means these are the diametric ints of circle. So, find the center and radius. So, simply we can write it as x minus 2 into x plus 2 plus y plus 3 means y minus y1, y minus y2 is equals to 0. This will be the required equation of circle. So, let's simplify it. It will be x square minus 4 plus y square minus 4 y plus 3 y that will be minus y minus 12 is equals to 0. Okay. So, finally, it becomes x square plus y square minus y minus 16 is equals to 0. Now, what is the center for this circle? Center will be no x term. So, center will be 0 comma 1 by 2, right? 0 comma 1 by 2 will be the center for this circle and what will be radius? Radius will be under root g square plus f square minus c. So, g square is 0. f square is 1 by 4 and minus c. Minus c means minus of minus 16 that will be 16. So, this will be 4 1 plus 64 under root that will be root of 65 by 2. Okay. So, this will be our answer. The center will be this and the radius will be root 65 by 2. So, 13th. Now, let's see question number 14. If 4 comma 1 be an extremity of a diameter of circle, find the coordinates of extremity of the other extremity of the diameter. Okay. So, it's saying let me consider it as a b, a b is the diameter of the circle. So, the points, coordinates of a is given as 4 comma 1. Okay. And we have to find the coordinates of b. Find the coordinates of the other extremity of the diameter. Okay. So, let me consider it as a comma b, but we know the equation of this circle. The equation of this circle is x square plus y square minus 2x plus 6y minus 15 is equals to 0. So, if equation of circle is known means, it means we know the center also. What will be the center? The center for this circle will be 1 comma minus 3. 1 comma minus 3, right? So, this 1 will be the coordinates of c will be 1 comma minus 3. Now, c will be the midpoint of a b, right? So, 4 plus a by 2 is equals to 1 and 1 plus b upon 2 is equals to minus 3. So, from here we get 4 plus a is equals to 2 or a is equals to 4 plus a is equals to 2. So, a is equals to 2 minus 4 means minus 2. And what will be b? b will be b plus 1 is equals to minus 6. So, b will be minus 7. So, this will be the coordinates of center, sorry, the other coordinates of b means coordinates of other extremity of diameter. Okay. Hope this is clear. So, let's move to the next question, question number 15. It is saying find the equation of circle drawn on the diagonal of the rectangle as its diameter whose sites are x equal to 4, x equal to minus 2, y equal to 5 and y equal to minus 2. Okay. Similar type of question we have done in this exercise only. I think in MCQ, but anyhow here we are not having options. We will do it once again, not an issue. So, this is what given find the equation of circle drawn on the diagonal. Okay diagonal. First, let me write it as x equal to minus 2. I am writing the equation of these lines x equal to minus 2 then x is equal to 4 and this line is y equal to minus 2 y equal to minus 2 and this is y equal to 5. Right. So, let me name it as A, B, C and D. So, coordinates of A will be, coordinates of A will be minus 2, x coordinate and y coordinate minus 2. What will be the coordinate of B? It will be 4 comma minus 2, coordinate of C will be 4 comma 5 and coordinate of D will be minus 2 comma 5. So, this will be the coordinate of D. Now, we have to draw one circle taking diagonal as the diameter, right? Find the equation of circle drawn on the diagonal of rectangle as its diameter. Yeah. So, it is saying that this is, this will be the diagonal of rectangle, right? So, considering this AC as diameter, considering this AC as diameter we have to draw one circle. Okay. So, this equation of this circle we have to find out, equation of this circle we need. So, we know the diametric and this AC will be nothing but diameter, no? This is the diameter of the circle and we know the coordinates of A and C. So, we can write it as x minus x1 means minus 2. That will become x plus 2, x minus x1, x minus x2 plus y minus y1. That will also become y plus 2 and y minus y2, y minus 5 is equals to 0. This will be the equation of the circle. Okay. Now, some people can say like I will consider this BD as the diameter. You can do that also. Considering BD as diameter, if you consider BD as diameter, right? So, what will be the equation of the circle? The equation of circle will be x minus 4, x minus x1, x minus x2 plus y minus y1 means x plus 2 and y minus y2, y minus 5 is equals to 0. So, if you observe, both these equations will be same, x minus 4 into x plus 2 plus y plus 2 into y minus 5. So, you can simplify this and get the answer. I am living here only. So, anyway, whether you consider AC as diameter or BD as diameter, you are going to have the same answer. So, this is question number 16. It is saying that the find the equation of circle which passes through point 1, 1, 2, 2 and whose radius is 4, 1. Okay. So, it's saying it is passing through 2 points. So, let me take it as A, 1, 1 and this is B, which is 2, 2 and we know the radius of this circle. We know the radius of this circle as 1. Okay. So, 3 informations are given. There's 2 points and 1 radius. Now, we have to derive the equation of the circle. So, let me write the equation of circle in the center radius form like this, x minus h whole square plus y minus k whole square is equal to r square. Where c is the center of this circle having coordinates as h comma k. We can write this standard form of circle in this way also where the centers in center radius form like h and k is the coordinate of center. And r is the radius of the circle. Now, this 1 comma 1 and 2 comma 2 must satisfy this equation. So, 1 minus h whole square plus 1 minus k whole square is equal to what is r? r is 1 means that is equal to 1. So, opening this, we get h square plus 1 minus 2h plus 1 plus k square minus 2k is equals to 1. Or h square plus k square minus 2h minus 2k minus 2h minus 2k plus 1 is equals to 0. Now, b should also satisfy this equation. So, putting that also 2 minus h whole square plus 2 minus k whole square is equals to 1. So, 4 plus h square minus 4h plus 4 plus k square minus 4k is equals to 1. That will be equal to 4h square plus k square minus 4h minus 4k plus 4 plus 4h minus 1 that will be 7 plus 7 is equals to 0. Now, comparing these two, this 1 and 2, they both represent the same thing. So, we can say this minus 2h minus 2k plus 1 is equals to minus 4h minus 4k plus 7. So, this will become 2h 4h minus 2h 2h 4k minus 2k plus 2k and 7 minus 1 that will be 6. So, h plus k is equals to 3. We got this relation h plus k equals to 3. Now, putting this in any of the equation, what we get? Let me put it in this equation. I will be getting 1. So, what will be h? H will be nothing but 3 minus k. So, putting h equal to 3 minus k, I will get 1 minus h, 1 minus h. So, 1 minus 3 plus k whole square plus 1 minus k square is equals to 1. So, it will give a quadratic, right? It will give a quadratic in k. Solving this, this becomes k minus 2 whole square plus 1 minus k square is equals to 1. So, this will be k square plus 4 minus 4k. Plus 1 plus k square minus 2k minus 1 is equals to 0. So, this will be 2k square minus 6k, 2k square minus 6k. This plus 1 minus 1 will be cancelled. Plus 4 equal to 0 or k square minus 3k plus 2 is equals to 0. So, k square minus 2k minus k plus 2 equal to 0. k, k minus 2 minus 1, k minus 2 equals to 0. From here, we get two values of k. k is equals to 1 or 2, okay? So, putting k equal to 1, if you say, when we put k equal to 1, what will our h be? Our h will become 3 minus 1, that is 2. And if when k is equals to 2, our h will become 1, okay? So, when k is 2, h is 2 and when k is 1, h is 2 and when k is 2, h is 1. So, putting this in our original equation like this x minus h whole square, this is what we assume the equation of circle. x minus h whole square plus y minus k whole square is equals to 1. So, let me put h is 2. So, this will become x minus 2 whole square plus y minus, when h is 2, k is 1. y minus 1 square is equal to 1. This will be our one first equation. And when we consider h as 1, I will get x minus 1 whole square. In that case, k will be 2. So, y minus 2 whole square is equals to 1. So, this is the second equation. So, both are our answer, right? So, this both will be our answer, okay? This will be also, this will be also our answer. So, moving ahead, this is our last question of this exercise, exercise number 2. So, it is saying that find the equation of circle, which passes through the points 3 comma 4, comma 3 comma minus 6 and 1 comma 2, okay? So, through three non-colonial points, only one end, one circle can pass, right? Through two points, infinite number of circles can pass, but through three non-colonial points, only one circle can pass. So, let me consider the equation of circle as our standard equation, that is x square plus y square plus 2 gx plus 2 fy plus c equals to 0. Let me consider this as the equation of the circle. Now, since it passes through all these points, it must satisfy all these, means all these three points must satisfy this equation. So, sorry. So, let's put it, it will be 9 plus, I am putting 3 comma 4. So, 9 plus 16 plus 2 into 3 6 g, 6 g plus 8 f plus c is equals to 0, right? So, further we can simplify it. So, this will be 6 g plus 8 f plus c equals to 69 minus 25. So, let me put it as equation one. And that second point will also satisfy it. So, this will be 9 plus 36 plus 2 g into 3, that is nothing but 6 g. And 2 y means minus 12 f, minus 12 f plus c is equals to 0. So, this will be 6 g minus 12 f plus c, 6 g minus 12 f plus c is equals to 36 and 9 is 45. 36 plus 10, 46, 45. So, that will be minus 45. So, let's consider it as equation two. And this third point will also satisfy this equation. So, 1 plus 4 plus 2 g, 2 f into 2, that will be 4 f, 4 f plus c is equals to 0. So, 2 g plus 4 f plus c is equals to 4 plus 1 is minus 5. So, this is our three equations and we are having g f and c unknowns, g f and c as unknowns and we are having three equations. So, we can solve it. So, let me subtract this second from first. We will get 8 f minus. So, this will become 20 f, c and c will be cancelled out. So, 20 f will be equal to minus 25 plus 45, right? So, 45 minus 25 will be 20. So, f will become 1, okay? f will be equal to 1 and we can subtract this also. Or let me simplify it. That will become 6 g minus 12 plus c equals to minus 45. Or 6 g plus c will be equals to minus 33. And this will become 2 g plus 4 plus c equals to minus 5. So, 2 g plus c equals to minus 9. So, 2 g plus c equals to minus 9. Now, subtract it. Subtracting we get 4 g. If you subtract it, what you will get? 6 g minus 2 g that will be 4 g minus 33 plus 9. That will be minus 24, right? So, g will be minus 6. And what will be c? If g will be minus 6, then it will be minus 12 plus 6 is equals to minus 9. That means c is equals to 12 minus 9, that is 3. So, our equation of the circle will become, our equation of the circle will become x square plus y square plus 2 into g, g is minus 6, 2 g x plus 2 into f. f is 1, 2 f y plus 3 is equals to 0. That is nothing but x square plus y square minus 12 x plus 2 y plus 3 equals to 0. This is the required equation of the circle. So, we are done with this exercise. We are done with this exercise. Okay. I think this exercise is clear to everyone. So, this is all for today. And we will meet soon with our next exercise, that is exercise number 3 of the circle. So, till then, take care. Tata, goodbye. Okay.