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  4. Consider the parabola x2=4 y and circle x2+(y-5)2=r2(r>0). Given that the circle touches the parabola at the points P and Q. Let R be the point of intersection of tangents to parabola at P and Q and S be the centre of circle. Tangents drawn from the point A (1,-1) to the circle x 2+ y 2-4 x +6 y -1=0 touch the circle at the points B and C. If the circumcircle of the triangle ABC cuts the auxiliary circle of hyperbola (- x 2/4)+( y 2/ b 2)=1 orthogonally, then b 2 is equal to

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Q. Consider the parabola $x^2=4 y$ and circle $x^2+(y-5)^2=r^2(r>0)$. Given that the circle touches the parabola at the points $P$ and $Q$. Let $R$ be the point of intersection of tangents to parabola at $P$ and $Q$ and $S$ be the centre of circle.
Tangents drawn from the point $A (1,-1)$ to the circle $x ^2+ y ^2-4 x +6 y -1=0$ touch the circle at the points $B$ and $C$. If the circumcircle of the $\triangle ABC$ cuts the auxiliary circle of hyperbola $\frac{- x ^2}{4}+\frac{ y ^2}{ b ^2}=1$ orthogonally, then $b ^2$ is equal to

Conic Sections

Solution:

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Clearly circumcircle of the $\triangle A B C$ is $(x-1)(x-2)+(y+1)(y+3)=0$ $\Rightarrow x^2+y^2-3 x+4 y+5=0$.....(1)
Also auxiliary circle of hyperbola $\frac{-x^2}{4}+\frac{y^2}{b^2}=1$ is $x^2+y^2=b^2 \ldots$ (2)
$\therefore$ By using condition of orthogonality, we get $b^2=5$