Equation of circles tangent to $x$-axis with a given radius and point Find the equations of the circles which are tangent to the x-axis, with radius of 5 units and passing through the point $(0,8)$.
I know how to formulate an equation with the given radius of 5 and then substitute the point $(0,8)$, but then I get stuck at finding their different centers.

The equation of circle is $(x-a)^2+(y-b)^2=r^2$ centered ar point $(a,b)$. 
The equation of line is $y=mx+c$. The x axis is the line $y=0$.
All circles having tangent to x axis implies it have point $y=0$ and only one solution for x.
$(x-a)^2+b^2=5^2$
$x^2-2ax+a^2+b^2-25=0$
We want solutions for x.
We thus require the discriminant to be zero. $b^2-4ac$ is 
$26-a^2-b^2=0$.
These are the equation which is to be solved for to find $a,b$ for single solutions.
$b^2=26-a^2$
Inserted into equation of circle with $y=0$
$x^2-2ax+a^2+26-a^2-25=0$
$x^2-2ax+1=0$
