Question

If the length of the tangent from $(5,4)$ to the circle $x^2+y^2+2ky=0$ is $1$, then find $k$

Answer 1

The radius of the circle, the tangent from $(5,4)$ and the point $(5,4)$ when joined to the center of the circle, $(0,-k)$ forms a right angled triangle.

$\therefore$ (length of tangent)${}^2$ + (radius)${}^2$ = (distance of $(5,4)$ from center of the circle)${}^2$

$\therefore$ (length of tangent)${}^2+k^2=(5-0{)}^2+(4+k{)}^2$

$\therefore 1+k^2=25+k^2+8k+16$

$\therefore 8k+40=0$

$\Rightarrow k=-5$