 Now associated with this concept of pair of tangents is director circle. So what is a director circle? Let's understand this concept. Director circle is the concept which says that the locus of all those points from where you can draw two such tangents. So locus of all those points from where you can draw two such tangents which are perpendicular to each other, then the points will trace the director circle, right? It's the locus of such points. So director circle is the locus of such points from where you can draw two tangents to the circle which are perpendicular to each other. So it's locus of the point, you can say locus of the point of intersection of perpendicular tangents to perpendicular tangents to a given circle, okay? How do I find the equation of that? Now guys, very, very simple, let's again use the fact that equation of a pair of tangents is t square equal to ss1, okay? So what will be t in this case? You will say xx1 plus yy1. So I'm taking the standard form of the equation of a circle, x square plus y square is equal to a square. So t square is equal to s. This is going to be s. This is going to be s1, okay? Now if this represents two tangents which are perpendicular to each other, always remember this one important thing, remember that if this represents a pair of straight lines which are made by such lines which are perpendicular to each other, so if this represents two perpendicular lines, if this represents two perpendicular lines, then a plus b will always be 0. So this plus this will always be 0. This plus this will always be 0. Just remember this, very, very important concept. I'll be proving it when I'm doing the straight lines, a pair of straight lines concept with you, but as of now just remember this concept. So now this is the combined equation of the pair of straight lines. So what is the coefficient of x square? Coefficient of x square you will say is nothing but x square on your left side and minus x square plus k square minus a square on the right side. So altogether it will be a square minus k square, okay? Similarly the coefficient of y square term would be what? The coefficient of y square term would be k square minus h square plus k square minus a square, which is going to be a square minus h square. And just now we discussed that coefficient of x square and y square should give you a 0, should give you a 0. It means a square minus k square plus a square minus h square should give you 0. That means h square plus k square should be equal to 2a square and if you generalize this you will get x square plus y square is equal to 2a square as the equation of the director circle. So this is the equation of the director circle, okay? Few pointers that you need to keep in mind. One is that you could have solved this equation very easily had you known the fact that the director circle is concentric with the director circle. So this circle is going to be concentric with the original circle. That means both will have the same center, okay? If that fact is known we can actually find the radius very very easily, okay? So this is going to be your radius of the director circle. This is a, this is going to be 45 degrees because each will be 45, 45. So this will be r will be root of 2a. So if you have to write the equation of any circle whose center is at origin and radius is root of 2a then you can easily write down the equation like this, okay? So that gives me the equation of the director circle as x square plus y square is equal to 2a square that we have already proved, okay? So few pointers guys that I want to discuss with respect to director circle. The first pointer being director circle is concentric with the given circle, with the given circle. That means they have the same center and secondly the radius of the director circle will be root 2 times the radius of the given circle. So remember these two facts you will always be able to write down the equation of a director circle very very easily. Any questions so far guys? Is it clear? Please type CLR on your screen if it is clear, alright. So we will move on to the last concept of the day which is the concept of chord of contact. What is chord of contact? Chord of contact is basically a concept which says that if you draw two tangents from an external point x1, y1, okay? The line joining the point of contact P and Q, this line is called the chord of contact. The line of the chord of contact is given by t equal to 0. That is let's say this circle is x square plus y square is equal to a square, then the chord of contact equation is given as this. It's very much similar to the equation of the, it's exactly the same as the equation of a tangent but note that this chord of contact will not be the tangent because your x1, y1 doesn't lie on the circle, it is outside the circle. So even though the representation of the equation is same, please note that the same representation is now representing something else, okay? How do we prove this? How do we prove this fact? Very simple. Let's say this point is alpha 1 beta 1 and this point is alpha 2 beta 2, okay? So if I ask you, write down the equation of tangent drawn at alpha 1 beta 1, you would say very simple, it will be x alpha 1, y beta 1 equal to a square, correct? Similarly if I ask you equation of tangent drawn at alpha 2 beta 2, you will say x alpha 2, y beta 2 equal to a square. And both these tangents, both 1 and 2 is satisfied by x1, y1 point. So as you can see here, this x1, y1 will pass through the tangent drawn at p also and tangent drawn at q also, isn't it? And if it satisfies both the, if both the tangent is satisfied by this point, I can say substitute x1, y1 in both 1 and 2. So if you substitute it in 1, you get x1 alpha 1, y1 beta 1 equal to, equal to a square and we get x1 alpha 2, y1 beta 2 equal to a square. Now guys, look at it very, very carefully, these two equations suggests that, it implies that alpha 1 beta 1 and alpha 2 beta 2 satisfy x1, x, y1, y equal to a square, isn't it? Because just replace your x with alpha 1 and y with beta 1, you will get the very first equation over here, clear? Similarly, if you replace your x with alpha 2 and y with beta 2, you will get the second equation over here, correct? So it, both of these suggest that alpha 1 beta 1 and alpha 2 beta 2 satisfy this equation and this is clearly the equation of a line. And the only line which is satisfied by both alpha 1 and alpha 1 beta 1 and alpha 2 beta 2 has to be the chord of contact, has to be the chord of contact. So see how dicey can be this proof is the chord of contact. So hence the chord of contact equation is going to be this expression, which is your expression written over here, t equal to 0. Now a very interesting thing, very, very interesting thing that you will observe over here is that if you start moving your x1, y1 closer to the circle, your chord of contact will start moving towards the periphery, right? So basically, as you shift it closer to the circle, you would realize that chord of contact will come closer and closer, okay? So if I put it over here, it will become like this. If I put it over here, it will become like this. So ultimately when this x1, y1 lands up on the circle, so when x1, y1 lands up on the circle, lands up on the circle, you realize that chord of contact will become a tangent at that point, will become the tangent, will become the tangent at that point. So see the beauty of the scene, that's why the same equation is being represented, the same equation is representing both the things, because the same chord of contact will end up becoming the tangent if this point x1, y1 happens to lie on the circle. Is that clear? Okay, so we'll end up this session by a quick problem. So last problem for the day. Question is, I'll draw it better, I'll better draw the question so that you can understand it in a simpler language. Okay, so these are two concentric circles, okay? This circle has a equation of x square plus y square, in fact, there's a third one also. So the outer one has a equation of x square plus y square is equal to a square. The second one has an equation of x square plus y square is equal to b square. And the third one has the equation x square plus y square is equal to c square, okay? Now guys, listen to this question very, very carefully. The question says, if the chord of contact, if the chord of contact drawn from, drawn from a point on the circle x square plus y square is equal to a square to the circle x square plus y square is equal to b square, touches the circle, touches the circle x square plus y square is equal to c square, then show that a, b, c are in GP, a, b, c are in GP. So basically what I want to convey over here is this, this is a scenario right now. So any point you have taken on, let's say the outer circle a, which forms a chord of contact on the second circle and this touches the circle. So it's a tangent at this point. So show that a, b, c are in, are in GP. Can you one minute to do this? Very simple question. Any idea guys? Anybody could do it. So let's discuss this. If I take this point to be a cos theta, a sin theta, okay? Then I'm drawing a chord of contact from a on to the second circle, correct? Okay. So I can write the equation of a chord of contact as t equal to 0. So x, x1 plus yy1 equal to b square. Remember you're drawing the chord of contact to this circle, correct? And the point is actually a cos theta, a sin theta. So instead of x1, you'll use a cos theta instead of y1, you'll use a sin theta, okay? So this will be your chord of contact bc. This is the equation of bc. Now this line is tangent to the inner circle, correct? If it is tangent to the inner circle, we have learned that the distance from the center of the circle to this point should be c, yes or no? So distance of origin, distance of origin from this line which I call it as 1 must be c, correct? So that would be mod b square by under root of a square cos square theta b square sin square theta equal to c, correct? Which is nothing but mod b square. b square is only positive so it doesn't need any mod for it. So b square divided by under root of a square, okay? a is also positive because it's the radius. So it's going to be this. So b square is going to be ac and hence ab and cr in gp. So this is how we can prove that this situation is going to be true, okay? So guys, thank you very much for coming online. So we did a lot of things today. I think I would need just one more class to wrap up circles. And then we can quickly go to parabola, ellipse and hyperbola in a similar way, okay? Thank you so much for coming online over and out from Centrum Academy here. Thank you so much for coming online over and out from Centrum Academy here.