Question

If the length of the tangent from $(2,5)$ to the circle $x^2+y^2-5x+4y+k=0$ is $\sqrt{37}$, then find $k$.

Answer 1

Center of circle =$(\frac{5}{2},-2)$

Radius of circle $=r=\frac{25}{4}+4-k$

As tangent is perpendicular to the radius, we apply Pythagoras theorem (see the figure)

$\Rightarrow A{O}^2=A{P}^2+P{O}^2$

$\Rightarrow \frac{1}{4}+49=r^2+37$

$\Rightarrow \frac{197}{4}=\frac{25}{4}+4-k+37$

$\Rightarrow k=13$