Question

If $y=2x$ is a chord of the circle $x^2+y^2-10=0$, find the equation of the circle with this chord as diameter.

Answer 1

The points where $y=2x$ meet on circle $x^2+y^2=10$ will be given by substituting the equation in the equation of circle.

$x^2+(2x{)}^2=10\Rightarrow 5x^2=10\Rightarrow x=\pm \sqrt{2}$

The points where the given line meets the circle are $(\sqrt{2},2\sqrt{2})$ and $(-\sqrt{2},-2\sqrt{2})$

The radius of the circle formed with above points as diametrically opposite points $=\frac{\sqrt{(\sqrt{2}+\sqrt{2}{)}^2+(2\sqrt{2}+2\sqrt{2}{)}^2}}{2}$

Radius of the new circle $=\frac{2\sqrt{10}}{2}\Rightarrow \sqrt{10}$

The center of the circle will be $=(\frac{\sqrt{2}-\sqrt{2}}{2},\frac{2\sqrt{2}-2\sqrt{2}}{2})=(0,0)$

Equation of the circle will be $(x-0{)}^2+(y-0{)}^2=(\sqrt{10}{)}^2\Rightarrow x^2+y^2=10$