Question

Consider the parabola $x^2=4y$ 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 $x^2+y^2-4x+6y-1=0$ touch the circle at the points $B$ and $C$. If the circumcircle of the $△ABC$ cuts the auxiliary circle of hyperbola $\frac{-x^2}{4}+\frac{y^2}{b^2}=1$ orthogonally, then $b^2$ is equal to

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

Answer: (D)

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