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  4. A circle C touches the line x=2 y at the point (2,1) and intersects the circle C1: x2+y2+2 y-5=0 at two points P and Q such that PQ is a diameter of C1 . Then the diameter of C is :

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Q. A circle $C$ touches the line $x=2 y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2 y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1} .$ Then the diameter of $C$ is :

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Solution:

$(x-2)^{2}+(y-1)^{2}+\lambda(x-2 y)=0$
$C: x^{2}+y^{2}+x(\lambda-4)+y(-2-2 \lambda)+5=0$
$C_{1}: x^{2}+y^{2}+2 y-5=0$
$S_{1}-S_{2}=0$
(Equation of $PQ$)
$(\lambda-4) x-(2 \lambda+4) y+10=0$
Passes through $(0,-1)$
$\Rightarrow \lambda=-7$
$C: x^{2}+y^{2}-11 x+12 y+5=0$
$=\frac{\sqrt{245}}{4}$
Diometer $=7 \sqrt{5}$