The equation of the circle is given by $x^2+(y-r)^2=r^2$ where $r$ is the radius. The point is located at the point $(d,h)$.
Here is my approach to this: the general equation of all lines that passes through the point can be written as $y=m(x-d)+h$, and the equation of the circle is $y=\pm\sqrt{r^2-x^2}+r$. So I equate them to each other and simplify to get a quadratic equation like this:
$$m(x-d)+h=\pm\sqrt{r^2-x^2}+r$$
$$(m^2+1)x^2+2m(h-r-md)x+(m^2d^2-2mdh+2mdr+h^2-2hr)=0$$
Since the line is tangent to the circle, I set the discriminant to zero and solve for $m$.
$$[2m(h-r-md)]^2-4(m^2+1)(m^2d^2-3mdh+2mdr+h^2-2hr)=0$$
This simplifies to:
$$(r^2-d^2)m^2+2d(h-r)m+(2hr-h^2)=0$$
Using the quadratic equation:
$$m=\frac{-d(h-r)\pm r\sqrt{d^2-2hr+h^2}}{r^2-d^2}$$
Now putting this back into the line equation $y=m(x-d)+h$, the x-intercept is found by setting y equal to zero, I found x to be:
$$x=d+\frac{h(r^2-d^2)}{d(h-r)\pm r\sqrt{d^2-2hr+r^2}}$$
I have verified this by plugging in values, but as you can see this is very messy, I am just wondering if there is a clean and easier way to do this.
For a circle $X^2+Y^2=R^2$ the equation of line tangent at $(X_1,Y_1)$ is given by $$(X_1)X+(Y_1)Y=R^2$$
In your question $$X=x,Y=y-r,R=r$$
Put the values and you get the equation of line from which X intercept can be found.