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A circle $\mathrm{C}$ touches the line $\mathrm{x}=2 \mathrm{y}$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2 y-5=0$ at two points $\mathrm{P}$ and $\mathrm{Q}$ such that $\mathrm{PQ}$ is a diameter of $\mathrm{C}_{1}$. Then the diameter of $\mathrm{C}$ is :
$7 \sqrt{5}$
$15$
$\sqrt{285}$
$4 \sqrt{15}$
Solution
$(\mathrm{x}-2)^{2}+(\mathrm{y}-1)^{2}+\lambda(\mathrm{x}-2 \mathrm{y})=0$
$\mathrm{C}: \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}(\lambda-4)+\mathrm{y}(-2-2 \lambda)+5=0$
$\mathrm{C}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{y}-5=0$
$\mathrm{~S}_{1}-\mathrm{S}_{2}=0$ (Equation of $\left.\mathrm{PQ}\right)$
$(\lambda-4) \mathrm{x}-(2 \lambda+4) \mathrm{y}+10=0$ Passes through $(0,-1)$
$\Rightarrow \lambda=-7$
$\mathrm{C}: \mathrm{x}^{2}+\mathrm{y}^{2}-11 \mathrm{x}+12 \mathrm{y}+5=0$
$=\frac{\sqrt{245}}{4}$
Diometer $=7 \sqrt{5}$
