A circle C touches the line x=2 y at the point (2,1) and intersects


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 :

  1. $7 \sqrt{5}$

  2. 15

  3. $\sqrt{285}$

  4. $4 \sqrt{15}$

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 $(0,-1)$

$\Rightarrow \lambda=-7$

$C: x^{2}+y^{2}-11 x+12 y+5=0$


Diometer $=7 \sqrt{5}$

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