 Hello, we are back once again with this exercise five of circle chapter. So we will be solving this exercise five of circles. Okay, so let's take the first question. It is saying if the tangent at P on the circle x square plus y square plus six six plus six y equal to two meets the straight line five x minus two y plus six equal to zero at a point q on the y axis. So, then the length of PQ. Okay. So, let me draw. Circle what is given there. So, here one circle is there was a question is known to us. Now one tangent at point P. So, I'm drawing one tangent at point P to the circle. That is meeting the straight line five x minus two y plus six. So one straight line is there. I don't know whether that straight line will intersect the circle or not. That is not important for me also, and I'm drawing one this y axis also right. So, this is the sketch as per question. So, this is our y axis. Let me reduce the thickness. So, this is basically our y axis, this green line. Okay. And what else. This is our line five six minus two y. This line is five six minus two y plus six equal to zero. This is point P where we have drawn one tangent that tangent is meeting this straight line on the y axis. This is what it is given and this is meeting it. This point is basically q. Okay. And this is a circle sequence is given as x square plus y square plus six x plus six y minus two equal to zero. This much information we are given in the question. Now, the question is asking to find the length of pq, we have to find pq length of pq right. This is what we need to find. So, okay, let's extract all the information from this equation of circle. So, the center of this circle is minus three comma minus three. The radius of this circle will be nine plus nine minus of two minus of minus two that will be plus two. That means under root 20. Right G square plus F square minus of C. So under root 20 under root 20 means four into five so two root five. So radius of the circle is two root five and center is minus three minus three. Okay. So, we need to find the length pq. Okay, can we identify the coordinates of point q. If you see the coordinates of point q. One thing is sure that since this is lying on the x y axis sorry, since it is lying on the y axis so x coordinate will be zero. Right. Now how to find the y coordinate put x equal to zero in this equation. So if you put x equal to zero in this equation. So put x equal to zero in this equation we are headed minus two y is equal to minus six, or we can say y equal to three. Okay. So the coordinates of q will be zero comma three. Now, q is a point outside the circle q is a point outside a circle and it is touching the circle at point p. Okay, so pq if you say me to find the length of pq pq is nothing but under root s one where s one means power of point q right power of point q under root is will be equal to length of tangent that is pq. Okay, so what will be s one s one will be x square plus y square plus six six plus six y minus two put x equal to zero and y equal to three. So, like, means we have to put the coordinates of q in this equation and we will, we will get this s one. So, if you say that x will be zero and this was zero plus y three y square will be nine plus zero plus six and two three, that is, six and two three know that will be 18 minus two. So, 27 minus two that is 25. So, if you see this length of pq length of pq will be equal to, or this is the length of tangent only this will be equal to under root s one means under root 25 that will be equal to five minutes. Okay. So, option C is correct. Option C is correct for discussion. Let's take the next question, question number two. So, what is saying if the circle x square plus y square plus two gx plus two a point plus equal to zero this is the general equation is given here is judged by y equal to x at p. Okay, such that OP is equal to six root to where is the origin, then the value of C. So one circle is here. This is our circle. Okay. And it is given it is touched by the line y equal to x. So, let me draw a coordinate axis. And let me draw this y equal to x line. So, it is touched by line y equal to x right. It is not cutting it is touching the line y equal to x. So, if you say, y equal to x will be somewhere. Okay, then he could have omitted this one. coordinate axis. Okay, let me draw it once again. Let me draw it separately. So, this is suppose this is our line y equal to x. Okay. And it is touching the circle. It is touching the circle. So, suppose this is our circle. And this is why y equal to x is touching this circle, such that OP at p. Okay. So, this is the line y equal to x. Sorry. This is the line y equal to x. This is our circle. And this line is touching this circle at point p. And one distance is there. OP is given as 6 root 2. So, this suppose I am taking this as whose coordinates are 0 comma 0. So, this OP is given as 6 root 2. OP is given as 6 root 2. So, this is our line y equal to x. So, this is our line y equal to x. OP is given as 6 root 2. So, what is OP basically? OP is nothing but the length of tangent. And we get the length of tangent as under root s1. So, this under root s1 means power of point under root. Right? So, power of which point, power of this origin. Under root of that will give us the length of OP. That is equal to 6 under root 2. Now, put the value 0 0 in the equation. We get 0 plus 0 plus 0 plus 0 plus c under root. That will be equal to 6 root 2. So, under root c is equal to 6 root 2. So, is squaring this, we get c. c is equal to 36 into 2. That is nothing but 70. Right? So, this will be our answer. The value of c will be 72 basically. Okay. Now, let's take the next question. Question number three. The God of contact of engines from a point P to a circle passes through Q. If L1 and L2 are the lengths of the tangents from P and Q to the circle, then P, Q is equal to. Okay, let's try to visualize it by trying a sketch. This is our circle. The God of contact from a point P. Suppose this is our point P from where we are growing the tangents. This is our pair of engines from point P. And this is our God of contact. Right? This is our God of contact, which is passing through Q, which is passing through Q. And if L1 and L2 are the lengths of tangents from P and Q to a circle. Okay. So, we have to draw tangents from Q also. Right? So, let me draw the tangents from Q. Suppose this is one tangent and this is another tangent. Okay. So, this is our point P. Basically, this is our point P. And this is our point Q. Okay. So, and this is our point of God of contact. Suppose this is AD passes through Q. If L1 and L2 are the length of tangents. So, this PA, this PA is equal to L1. Right? This PA is equal to L1. And this Q, let me name it as CD. Let me take it as a point C. And this has point D where this tangents are touching the circle. So, PA is given to be L1. And this QC is given to be L2. Right? So, this is what it is given. And we need to find the length of PQ. Then PQ we need to find this length PQ. PQ we need to find. This is not equal to us. So, if you say, let's assume the coordinates of P is X1 comma Y1. And the coordinates of PQ is X2 comma Y2. Right? So, what will be PA basically? PA will be equal to under root S1. Means power of point, power of point P. And taking under root of that will give me the length PA. So, this will be equal to... Okay, let's me... First, we have to assume the equation of circle. So, let me assume this circle as X square. Let the equation of circle as X square plus Y square is equal to A square. Okay? We are assuming this circle. So, this S1, under root S1 will be under root of X1 square plus Y1 square minus of A square. For power of points, we put the coordinates in the equation of the circle itself. So, this is what we have done here. And what will be our QC? This will be under root S, power of P I am telling. So, this is I am telling as power of Q. Okay? That will be nothing but X2 square plus Y2 square minus of A square. Okay? And what will be PQ? If you see the PQ, PQ is nothing but... PQ is nothing but distance between... We can write it the distance between these two points. So, this is nothing but X1 minus X2 whole square. Okay? Plus Y1 minus Y2 whole square. Okay? So, let's open it. It will become X1 square plus X2 square plus Y1 square plus Y2 square and minus 2 X1, X2 from here and minus 2 Y1, Y2. Right? Whole thing will be under root. Okay? One more thing is given here. If you observe, let me... So, if you see, this is the equation of AB, like equation of chord of contact we can write. Equation of chord of contact. How we used to write the equation of chord of contact? We used to write it as T equal to zero. Now, T means X, X1 plus Y, Y1 minus A square is equal to zero. So, X1 and Y1, I'm taking the coordinates of P. Like from P, I have drawn a pair of tangents. Okay? That touches the circle at A and B. So, this equation of AB, the equation of the chord of contact is given by T equal to zero. Where T is nothing but we replace the equation of circle, means in the equation of circle we replace X square by X, X1, Y square by Y1. So, this is what we have done. And so, this is our equation of AB. Right? This is nothing but equation of AB. This is our equation of AB. But this is passing through Q also. So, this must satisfy Q. It must satisfy Q. So, this will be, in place of X, we can write X2. In place of Y, we can write Y. So, this will be X2 into X1 plus Y2 into Y1 minus A square is equals to zero. Okay? So, we will use this information here. How? If you see, this will become X1 square plus X2 is equal to plus Y1 square plus Y2 is square, taking two common, minus two common, we are having X1, X2 plus Y1 by two. Right? So, in place of this X1, X2 and Y1 by Y2, what I will do? I will write it as minus of A square. I will write it as minus of A square. Right? Sorry, A square, not minus of, because after putting it to the right side, it will become plus A square only. So, here we get this X1 square plus Y1 square and minus two A square, I will split it into minus of A square. Okay? And plus, this will be Y1 square, Y1 is already, we have to, this will be Y2 is square and X2 is square. This X2 is square plus Y2 is square and minus of A square. Okay? So, this is nothing but, what is X1 square plus Y1 square minus A square? This is equal to L1 square. Okay? So, from here you can see, this is L1 and this is L2. So, squaring this, we are having this L1 square plus L2 square. Okay? So, this will be the length of P cube. Under root L1 square plus L2 square. Right? So, option C is correct. Option C is correct. This question. Now, let's take the next question, question number four. It is saying, if a chord of contact, if the chord of contact of tangents from a point X1 comma Y1 to the circle this touches the circle X minus A square plus Y square equal to A square, then the locus of X1 and Y1. Okay? So, here it is given that this is the circle. This is one circle. And we had one pair of tangents is drawn from this point Q, from this point P. Okay? So, from this point, a pair of tangents is drawn and we are now making this is the equation of, means this is the chord of contact, which when extended touches the circle, this one. Okay? So, one more circle we have to draw there. One more circle is there. Okay? So, this is touching this chord of context is touching this circle. So, if you see, this is our point P, suppose I'm taking this point as P. Okay? From a point X1 Y1, which coordinates are basically X1 and Y1. So, from this point of pair of tangents is drawn. And this is our part of contact, this AB. AB is our part of contact. And this, what is the circle? This circle is X square plus Y square is equal to A square. So, AB, when extended or AB when produced, it touches the circle, this one. This circle is nothing but this circle is X minus A whole square and plus Y square is equal to A square. Right? So, this is what it is given. So, what do we need to find? We need to find the locus of this P. We need to find this locus of P. So, how can we calculate it? We need to find the locus of P. Okay, we can draw or we can say P. What can we do? Whether writing the equation of AB will be helpful. Let me try. Let me write the equation of AB. The equation of AB will be basically T. Equation of AB will be T equal to 0. Where X, X1 plus Y, Y1 minus A square is equals to 0. Okay, I have written the equation of AB. I have written the equation of AB. This will be our equation of AB. Now, this is the straight line and okay. This is, what is the center of the second circle? This is A comma X minus A whole square plus Y minus 0 whole square. So, this A comma 0. This is the center of this circle. Okay, so if you see, if you draw the perpendicular from this C on this part of contact, it will be 90 degrees. So, let me take this point as C, P, Q. Okay, let me take as Q. So, this C, Q will be equal to radius. C, Q will be radius of this circle. Radius of this circle is equal to A, right? So, this C, Q must be equal to A. Now, how to get this C, Q? So, it will be A into this means X1A, right? Plus 0 minus A square upon under root of X1 square plus Y1 square. And this must be equal to A, right? So, taking this in model. So, further on, further simplification if you see, this will become, if I take A comma, it will become X1 minus A is equal to A into X1 square plus Y1 square. So, we can cancel this A. And we can take the square. We can take a square on both sides. So, it will be X1 square plus A square minus 2AX1 is equal to X1 square plus Y1 square. So, this X1 square, this X1 square will be get cancelled out. And we are left with Y1 square is equal to minus 2AX1 plus A square, right? So, this is the, like, this is what we got in terms of X1 and Y1. For low cost, for low cost, what we do? We used to replace this X1 and Y1 by X and K. So, our equation finally becomes Y square equal to minus 2AX plus any constant A square. Okay. So, this is nothing but, if you observe, this is nothing but the equation of a parabola. Y square is equal to 4X. It is of the same form or minus 4X. So, this is a parabola. Means the equation, means the locus of this point P is coming out to be a parabola. So, this will be our answer. So, option B is correct. So, this is clear to all. So, let's move to the next question. Question number 5. It is saying the locus of the midpoints of a part of the circle X square plus Y square equal to 4, which subtends a right angle at the origin. Okay. So, one circle is there. One circle is there. And one chord is there. Okay. So, let me draw this chord. And it is subtending. This chord is subtending the right angle at the origin. Right angle at the origin. Okay. So, the center is the origin for this circle. This is subtending 90 degree at the origin. Okay. So, let me name it. Let me name it as O, which is the origin and which is the center of the circle. The coordinates are 0 comma 0. Let me call, name it this chord as AB. And this angle AOB is given as 90 degree. This is what it is given in the question itself. And the equation of circle is known to us. This is X square plus Y square is equal to 4. Means center is at origin and radius is equal to 2. Radius is equal to 2 for this circle. So, this AO will be equal to 2. This OB will also be equal to 2 since this is the radius only. We have to do one more construction. Like what I will do? I will drop a perpendicular from this O on this AB. Okay. So, this will be 90 degree and this will bisect this. This will bisect this AB. So, let me name it as C. So, this AC must be equal to CB. Now, what we need to find? We need to find the locus of the midpoint. Okay. So, this is the midpoint of the chord. This C is the midpoint of the chord. Let me assume it as coordinates as H comma K. We need to find the locus of the midpoint of the chord, this chord, which is up to 90 degree at the origin. So, if you observe, if you observe the triangle, the bigger triangle, this triangle AOB. If you observe triangle AOB. So, this is a right angle triangle basically where this 2 is square plus 2 is square is equal to AB square, right? So, our AB is square is equal to 4 plus 4, 8. Or from here, we can say AB is equal to 4 into 2, means 2 root 2, right? If you observe in the bigger triangle AOB. Now, let's take the smaller triangle. Which triangle? Anyone? OCB. Let's take a triangle OCB. Triangle OCB. I'm taking triangle OCB. Here also, if you see, this is also a right angle triangle. So, here OC is square. OC is square plus BC is square. This must be equal to radius square, that is 2 square. Okay. And what is unknown here? OC, we don't know OC. BC, we know. BC is half of this AB. Okay. So, let me write it separately. What will be OC? OC is the distance. Okay. So, we can write here. OC is H minus 0 square H square plus K minus 0 square means K square under root. But it is square so that under root will be cancelled out. So, this will be H square plus K square plus BC square. So, this will be BC. BC will be half of AB, right? Half of AB. BC will be half of AB. That is nothing but root 2. So, root 2 guys square, that will be 2. And that is equal to 2 guys square. That will be 4. So, now replace this H and K by X and Y. So, we get the locus of the midpoint of this part as X square plus Y square is equals to 2. Right? This is the locus of the midpoint of the part which is subtening 90 degree at the center. So, X square plus Y square is equal to 2. So, option D is correct. Option D is correct for this portion. Now, let's take this next question, question number 6. It is saying the length of the tangents from P whose coordinates are given Q. Coordinates are given to a circle R root 2 and root 6 respectively then the length of the tangent from R. So, we know length of the tangent is nothing but the under root of the power of certain. This is what we know. Length of tangent is under root of power of that point from where the tangent is drawn. So, from P if you see, if you see from P. So, I am writing it as under root of power of point P that is equal to what will be there. So, let me assume the equation of circle, let the equation of circle, let the circle be, let the circle be. General circle that is X square plus Y square plus 2 g X plus 2 f Y plus equal to 0. So, we will apply here. So, under root of means power of point P under root will be given as what is the coordinate of P. That is 1 comma minus 1. So, this will be 1 plus 1 plus 2 g plus 2 g minus 2 f and plus C and that is given out to be a root 2. Okay. So, from here if you see, we are getting 2 plus 2 g minus 2 f plus C is equal to 2, right. So, this 2 2 will be cancelled. So, 2 g minus 2 f plus equal to 0. This is one equation what we got. Okay. Now, we know the power of this Q also means length of tangent from Q also. That is nothing but under root of power of Q point that will be, we know the coordinates of this Q. So, this will be a 9 plus 9 plus 6 g, right, and plus 6 f plus 6 f plus C that is equal to root 6. So, this is 18 plus 6 f plus 6 g plus 6 f plus C is equal to 6, right, is square in both sides. So, finally, we get 6 g plus 6 f plus C is equals to minus 12. So, this is our second equation. Now, we need to find the length of tangent from R whose coordinates are minus 2 comma minus 7. So, this is what we need to find, length of tangent from point R. So, this will be nothing but under root of minus 2 square like this will be 4 plus 49 and minus 2 into 2 g that will be minus 4 g. Then, 7 to 14 like that will be minus 14 f, right, minus 14 f plus C. So, this is we need to find out. This is what is asked in the question. So, from equation 1 and 2 if you see, so if you see if you multiply equation 1 by 3, equation 1 by 3 and then plus equation 2. I'm multiplying equation 1 by 3 and I'm adding with equation to this f term will be cancelled out. So, this will be 6 g, 6 g minus 6 f plus 3 C equal to 0. And now I will add it with equation 2. I will have them 6 g plus 6 f plus C equal to 0. So, this minus 6 f plus 6 f got cancelled out. This is equal to 12 g and plus, okay, minus 12 is also there in the equation 2. So, this is not equal to 0. This is equal to minus 12, right. So, this is 12 g plus 4 C equal to minus 12. Okay, 6 g, 6 g, 12 g plus 4 C equal to minus 12. So, 4 if you take 4 common it will be 3 g plus C equal to minus 3. So, from here we get g as minus 3, minus C equals 3, right. This is what we got the value of g here. Now, similarly, we find the value of f also in terms of C. So, put it in the equation 1, this value of g. So, 2 into g is minus 3, minus C upon 3, 2 g minus 2 f plus C equal to 0. Okay, so from here we get minus 2, 3 into 3 minus 1. So, minus 2, minus 2 C upon 3 minus 2 f plus equal to 0. This is 3, this is minus 6, minus 2 C plus 3 C is equals to 2 f, right. So, C, 3 C minus 2 C minus 6 upon 6 is equal to f. So, this is what we got f, means we find the value of g and f in terms of C. Now, I will put the value of this g and f here. So, putting this we get under root of sr is equal to under root of 4 plus 49 that is 53 minus 4 into g. g is minus 3 minus C upon 3 minus 14 into f, f is c minus 6 upon 6 plus C, right. So, this will be 53 minus 3 upon 3 means that will be plus 4, right. Minus 3 upon 3 minus 1 into 4 then minus C by 3 that will be plus 4 C by 3 and here we get minus 7 C by 3 plus 14. Plus 14 and plus C. So, here you see 4 C by 3 plus C will be 7 C by 3 and here minus 7 C by 3 this whole thing got cancelled out. So, we got 53 plus 4, 57, 67, 71. So, this power of point with respect to r or we can say length of tangent from r is equal to root of 71. So, this is our answer. So, option b is correct. Option b is correct for this. Let's move to the next question, question number 7. It is saying if the angle between the tangents drawn to x square plus y square plus 2 gx plus 2 y plus equal to 0 from 0 comma 0 is pi by 2. Okay, angle between the tangents. So, it is like fine tangent first, then I will put one circle here. So, if the angle between the tangents drawn to this is our circle, general circle only x square plus y square plus 2 gx plus 2 f y plus equal to 0 from 0 comma 0. So, this is our origin we are drawing. So, we have made pair of tangent from origin whose coordinates are 0 comma 0 is pi by 2 this angle is given to be this angle is given to be pi by 2, right. So, what we have to identify, then we have to comment on this g square plus f square. Okay, this angle is pi by 2, right. So, we can do one thing we can make we can we can do one construction, what is that construction. So, basically this is our center of the circle, okay, whose coordinates are minus g comma minus f, because it is a general circle only. And this angle will be basically 90 degree and this angle will be pi by 4, because this OC will, OC will bisect this angle with the pair of tangents. So, if you observe here, 10 pi by 4 means 10 pi by 4 will be equal to I am taking this point as P. So, this will be basically PC upon OP, okay, this will be PC upon OP 10 pi by 4 is nothing but one so this PC will be equal to OP. Now, what is this OP, this OP is the length of tangent from length of tangent. Okay, and what is this PC this PC is nothing but radius of the circle. So, we can write here, what is the radius for the circle this is under root of f square g square plus f square minus C, and that will be equal to under root means power of power of this point or is it with respect to the circle. So, all thing will be all the time will be zero only we are left with this C. Now is firing this, we got g square plus f square minus C is equal to C, or we can say g square plus a square is equal to 2C. Right, so this is our required answer g square plus f square is equal to 2C. So, this option B is correct. Take this question number eight. It is saying the course of contact of the pair of tensions drawn from each point on the line this to the circle this passes through a fixed point. Okay. Maybe confusing but let me draw it it will be. It will be for you to understand. So, this is the circle. Now I will draw a pair of tangents. Okay, a pair of tangents from this point. This point. This point is line one this line right and we need to we have to construct this point point of corner sorry corner of contact also. Okay, so let me draw this part of contact also. This is what this is what we need to draw. Now let me stay here. So, what is given the God of contact. Let me start from day. This is point B outside the circle. Okay, from here we have drawn the drawn a pair of tangents. And this is this AB is the part of contact. Okay, this is the center of the circle, which is origin basically 0, 0 and the radius of this circle is equal to one. So, this is known to us. Okay, one important thing is that this point P is lying on this line, whose equation is 2x plus y equal to four. Okay, this point P is lying on this line segment 2x plus y equal to four or we can say 2x plus y minus four equal to zero. So, suppose if I'm taking the coordinates of P as H, the x coordinate as H, then what will be its y coordinate, y coordinate will be four minus two H, right. The y coordinate will be four minus two H. Okay, now this, it is given that this AB, this God of contact always pass through a fixed point. Actually, this point P will vary, right? This point P can be anywhere on this line. If that is the condition, this God of contact will definitely pass through a fixed point and we need to identify that. Okay, so let me first write the equation of God of contact. Equation of AB, let me write it first. So, what will be the equation of AB? It will be t equal to zero. That means xx one. What is the equation of circle? The circle is x square plus y square is equal to one. Right, so xx one plus y y one minus one equal to zero. Now put x one and y one as the coordinates. Now what is x one? x one is h plus y and y one, y one, we can replace it by four minus two h minus one equal to zero. So, this finally becomes hx plus four y minus two h y minus one equal to zero. Now I will write it in this way. This, what you say, this four y minus one, I will group it in this way. I am grouping, taking these two terms and I am taking h common from these two terms. So, it will be x minus two y equal to zero. Actually, if you observe this, this is nothing but a family of the lines. Means this is becoming of, this is like L one plus lambda times L two four. Right, this is representing this formula. This is the equation of line, this L one plus lambda L two. So, all the lines, like passing through the intersection of this L one equal to zero and L two equal to zero means, what is this basically? This is basically a family of lines, okay, which passes through the intersection of L one equal to zero and L two equal to zero. So, it is clear now, like what is this represent? This represent a family of lines, okay, which passes through the intersection of L one equal to zero and L two equal to zero. So, what is L one here? If you see L one, so this will be four y minus one equal to zero. So, from here we get y equal to one by four. And what is L two here? L two here is x minus two y equal to zero means x equal to two y. So, two into one by four that is one by two. So, x is coming out to be one by four. So, this family of line will always pass through the fixed point that is one comma two and one comma four. Okay, so this will be the fixed point that is fixed point one comma two comma one by four. So, this will definitely pass this chord of contacts will definitely pass through this fixed point. If we change the position of P also, this will definitely pass through this fixed point. So, option number C is correct. Option C is correct for this question. So, I am stopping this exercise here only like we are done up to this question number eight. We will take the, we will take further questions in the next video. Okay, we will, I will try to record it as soon as possible. So, till then Tata and take care. So, I will release one more video for this exercise this exercise five part two. Okay, so there we will cover the remaining questions of this exercise. So, thank you. Thank you all of you.