A circle C touches the line x=2y at the point (2,1) and intersects the circle C_1:x^2+y^2+…

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

Accepted Answer

$C:(x-2)^2+(y-1)^2+\lambda(x-2y)=0\Rightarrow x^2+y^2+(\lambda-4)x-(2+2\lambda)y+5=0$.

Common chord $PQ$: $C-C_1=0$: $x(\lambda-4)-y(4+2\lambda)+10=0$. Since $PQ$ is a diameter of $C_1$, it passes through $C_1$'s centre $(0,-1)$:
$$-1\cdot(-4-2\lambda)+10=0\Rightarrow4+2\lambda+10=0\Rightarrow\lambda=-7.$$

$C:x^2+y^2-11x+12y+5=0$; radius $=\sqrt{\left(\frac{11}2\right)^2+36-5}=\sqrt{\frac{245}4}=\frac{\sqrt{245}}2$.

$$\text{Diameter}=\sqrt{245}=7\sqrt5.$$

Solutions · 1