Question

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

• A. $\sqrt{285}$
• B. $15$
• C. $4\sqrt{15}$
• D. $7\sqrt{5}$

Answer 1

The correct option is D $7\sqrt{5}$


Family of circles$\left(C\right)$ touches the line $x=2y$, at point $(2,1)$
$C:(x-2{)}^2+(y-1{)}^2+\lambda (x-2y)=0\cdots \left(i\right)$
$C_1:x^2+y^2+2y-5=0$ has cemtre $(0,-1).$

Common chord $PQ$ is
$PQ:C-C_1=0$
$\Rightarrow PQ:(x-2{)}^2+(y-1{)}^2+\lambda (x-2y)-x^2-y^2-2y+5=0$
$\Rightarrow PQ:x(\lambda -4)+y(-2\lambda -4)+10=0$
$\because (0,-1)$ lies on $PQ,$ then
$\Rightarrow 0+(2\lambda +4)+10=0$
$\Rightarrow \lambda =-7$

Putting $\lambda =-7$ in equation $\left(i\right)$
$(x-2{)}^2+(y-1{)}^2-7(x-2y)=0$
$\Rightarrow x^2+y^2-11x+12y+5=0$
$r=\sqrt{\frac{121}{4}+36-5}=\sqrt{\frac{245}{4}}=\frac{\sqrt{245}}{2}$

Diameter of $C,d=2r=\sqrt{245}=7\sqrt{5}$