Question

The line $2x-y+1=0$ is tangent to the circle at the point $(2,5)$ and the center of the circle lies on $x-2y=4$. Then find the radius of the circle.

Answer 1

Y = t satisfy equation of circle $x–2y=4$, $x=4+2\left(t\right)$ $\Rightarrow C(4+2t,t)$

Distance formula $r=\left(\left|\frac{a{x}_t+by+c}{\sqrt{a^2+b^2}}\right|\right)=\left|\frac{2(4+2t)-t+1}{\sqrt{5}}\right|=\left|\frac{3t+9}{\sqrt{5}}\right|$

$Radius=\sqrt{{\left(4+2t-2\right)}^2+{\left(t-5\right)}^2}$

$\sqrt{{\left(2-2t\right)}^2+{\left(t-5\right)}^2}$

$\therefore \frac{\left|3t+9\right|}{\sqrt{5}}=\sqrt{{(2-2t)}^2+{(t-5)}^2}$

$9t^2+54t+81=5(5t^2-2t+29)$

$t^2-4t+4=0$

$(t-2{)}^2=0$

$t=2$

$C=(4+2t+t)=(8,2),radius=\sqrt{5{\left(2\right)}^2-2\left(2\right)+29}=3\sqrt{5}$