Find the parameter $k$ given the equation of a circle

Question

Find the parameter $k$ so that the circle $$x^2 +y^2-(5k-1)x + (4-2k)y =5k$$ touches the $x$-axis. I know that if $C(p,q)$ is the center of the circle, then $q=r$ because it touches the $x$-axis. The answer is $k=-\frac{1}{5}$ but I don't know how to solve. Help please!!

Answer 1

when y=0 for x-axis

$$x^2-(5k-1)x -5k=0 $$

Determinant of quadratic equation is zero for tangent or double / coincident contact point at the time of intersection with x-axis.

$$(5k-1)^2-4 \cdot 1\cdot -5k =0$$ simplify $$(5k+1)^2=0,\quad k= -\frac15 , -\frac15. $$

Answer 2

Complete the squares, so that it looks like $(x-x_0)^2+(y-y_0)^2=r^2$. As a result $x_0,y_0,r$ will be functions of $k$ Solve for $k$ using $y_0=r$.

$x_0=(5k-1)/2$ and $y_0=k-2$, so $r^2=x_0^2+y_0^2+5k$. $y_0^2=r^2$ leads to $x_0^2+5k=0$ or $k=-\frac{1}{5}$.