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 $P Q$ is a diameter of $\mathrm{C}_{1}$. Then the diameter of $\mathrm{C}$ is :
Correct Option: 1
$(x-2)^{2}+(y-1)^{2}+\lambda(x-2 y)=0$
$\mathrm{C}: \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}(\lambda-4)+\mathrm{y}(-2-2 \lambda)+5=0$
$\mathrm{C}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{y}-5=0$
$S_{1}-S_{2}=0$ (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 $=7 \sqrt{5}$
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