 Hello students welcome to Centrum Academy Roup channel so today we have got for you a very very interesting question which is actually a good combination of differential equation and coordinate geometry. I'm sure you would love this question so let's read this out. Find the foci of the conic passing through the point 1 comma 0 and satisfying the differential equation 1 plus x squared times dx minus xy times dy equal to 0. Find also the equation of the circle touching the conic at root 2 comma 1 and passing through one of its foci. All right so let's solve the first part of the question we would like to find out what is this conic which is actually satisfying this differential equation. So this is a simple differential equation which is actually a variable separable form so we'll be trying out the separation of the variable in this case by separating the x with dx and y terms with dy so you can see it's easily variable separable. So let's now integrate both sides of the differential equation so the left hand side becomes ln mod x whereas the right hand side becomes half ln mod 1 plus y squared. Okay and of course let's not forget to put a constant of integration after performing the integration on both sides. Now how to find this constant of integration? We have been provided that this particular conic passes through the point 1 comma 0 it passes through the point 1 comma 0 so let us use this information to get the value of the arbitrary constant. So when we do that we get ln 1 because x value is 1 and half mod 1 plus 0 square because y value is given to us as 0 which brings us to the value of c as 0 itself. Right everybody knows ln mod 1 is ln 1 which is 0 so using the information we got the value of c as 0. So finally the differential equation solution finally becomes ln mod x is equal to half ln mod 1 plus y square which is nothing but 2 ln mod x equal to ln 1 plus y square which is finally nothing but ln of x square is equal to ln 1 plus y square and we don't need a mod around 1 plus y square because that's an already positive quantity. So removing the log terms on both the sides so it brings us to the conic section equation which is x square minus y square is equal to 1. So those students who have already done hyperbola in your class 11 would easily recognize this conic to be a rectangular hyperbola. Okay so the conic given to us finally comes out to be a rectangular hyperbola. Let's read the further parts of the question. Now we have to find the equation of a circle which touches this conic at root 2 comma 1 and passes through one of its foci. So it's very important to first find out the foci of this conic. So we all know that for a hyperbola the foci are at ae comma 0 and minus ae comma 0. Okay so this is a standard case of a hyperbola so for the standard cases we know that foci is at ae comma 0 and minus ae comma 0. A here is clearly a 1 and e value is a root 2. We know that for a rectangular hyperbola the eccentricity value is root 2. So the foci coordinates are finally root 2 comma 0 and minus root 2 comma 0. So now let's move forward to get the equation of the circle. So for equation of a circle let me sketch a small diagram for you all. All right dear students so basically we need a circle which passes through these two points and touches the hyperbola at root 2 comma 1. Let's call that point to be point p. So how do we solve this kind of a question? This actually is a case of a family of circles. A family of circles touching a given line at a given point. Let's say the point is alpha comma beta and the line equation is l equal to 0 is given by this equation x minus alpha whole square y minus beta whole square plus lambda l equal to 0. So this is the equation of the family of circles passing through alpha comma beta point and also touching the line l and this alpha comma beta point is lying on that particular tangent. So I'll be using a similar situation in this case as well where my equation of the line is the equation of the tangent drawn to the hyperbola at root 2 comma 1. So let's find out what is the equation of the tangent l equal to 0 at root 2 comma 1. So for that I'll be using my t equal to 0 formula which is xx1 minus yy1 equal to 0. So in this case I'll end up getting root 2x minus y equal to 1. And the point of course alpha beta is root 2 comma 1 for me. So using this family of circles we'll find out the equation of the required circle. So equation of the required circle would be certainly x minus root 2 whole square y minus 1 square plus lambda times root 2x minus y minus 1 equal to 0. Now the value of lambda is required to get this equation of the circle. So how will I get the value of lambda? I haven't just still used the fact that the circle has to pass through the focus root 2 comma 0. So this is one of the four sides. So I have to use this information to get the value of lambda. So what I'm going to do is root 2 comma 0 must satisfy the above equation. So if I do that I end up getting root 2 minus root 2 square 0 minus 1 square plus lambda and this expression will become root 2 into root 2 minus 0 minus 1 equal to 0. So let's simplify this further. So this is going to give me 1 plus lambda twice minus 1 equal to 0. So lambda value comes out to be minus 1. Okay so finally putting this value back let's see what is the equation turning out to be. So it's going to be x minus root 2 square y minus 1 the whole square minus 1 times root 2x minus y minus 1 equal to 0. Now this is one of the equations of the circles that we need. Mind you the question setter has said that it is passing through one of the four sides. Now this four side could also be the other four side whose coordinates would be negative root 2 comma 0. So if I use that information let us see what is the equation of the circle coming out to be. So I'll go back to my equation of the family of circles which is this and here I'll be putting the coordinates to be negative root 2 comma 0. So let's redo this entire process once again. So x minus root 2 square y minus 1 square plus lambda times root 2x minus y minus 1 equal to 0 and let's make the four side let's make the four side negative root 2 comma 0 satisfy the above equation. All right so when I do that I end up getting negative root 2 minus negative root 2 square 0 minus 1 square plus lambda times root 2 times negative root 2 minus 0 minus 1 equal to 0. So on simplification this gives us 8 plus 1 plus lambda minus 3 equal to 0. So lambda value comes out to be 3 in this case so putting this back in the equation of the family of circles over here the required family of circles that we get is this x minus root 2 whole square y minus 1 the whole square plus twice of root 2x minus 1 minus 1 equal to 0. So this is my another circle which satisfies the same set of conditions. So I hope dear students with this problem you are able to understand few concepts of differential equation as well as the concept of family of circles which is a very very important concept in JEE advance and JEE main exam. Thank you so much for watching. Please do like, subscribe and comment.