JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
A circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_1:x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_1$. Then the diameter of $C$ is :
Circle
Solution
$C$: $(x-2)^2+(y-1)^2+\lambda(x-2y)=0$ (touches the line at $(2,1)$). $C_1$: centre $(0,-1)$, $r_1^2=6$. PQ is a diameter of $C_1$, so the common chord $C-C_1=0$ passes through $(0,-1)$: $4+2\lambda+10=0\Rightarrow\lambda=-7$. $C: x^2+y^2-11x+12y+5=0$, $r^2=\tfrac{121}{4}+36-5=\tfrac{245}{4}$. Diameter $=\sqrt{245}=7\sqrt5$. Option (D).