Q:

Prove that center of circle and point of intersection of circle and x axis will have same X-coordinate.

My question is:

How can I prove if a circle, which have center say $(c,0)$,
touching (not cutting) X-Axis at some point will have same X-coordinate of center and X-Axis: $c$?

Let me explain the question. You may know that equation of a circle passing through X-Axis is given by:
$$
x^{2}+y^{2}-2 h x-2 a y+h^{2}=0
$$
Where $h$ are coordinate of circle and $a$ is radius.
Derivation of this general equation is given as following:

Consider a point $C(h,k)$ as center of a circle and touching the
X-Axis at point $M$.

Diagram:
!

Now $M$ has Y-coordinate 0 since it is on X-Axis. And X-coordinate $h$
(why? this is my question)
So now we have two points $C(h,k)$ and $M(h,0)$. The distance between
these two points will be radius (say $a$) of the circle.
So  $$a^{2}=(h-h)^{2}+(0-k)^{2}$$ $$\Rightarrow a^{2}=k^{2}$$
$$\Rightarrow \quad a=k$$
Hence on putting values in general equation of circle ie,
\begin{align} (x-h)^{2}+(y-k)^{2}=r^{2} \end{align}
We get: $$ \begin{aligned} &\Longrightarrow x^{2}+h^{2}-2 h
 x+y^{2}+a^{2}-2 a y=a^{2}\\ &\Longrightarrow x^{2}+y^{2}-2 h x-2 a
y+h^{2}+a^{2}-a^{2}=0\\ &\Longrightarrow x^{2}+y^{2}-2 h x-2 a y+h^{2}+a^{2}-a^{2}=0\\ &\Longrightarrow x^{2}+y^{2}-2 h x-2 a
y+h^{2}=0 \end{aligned} $$

My question is How can we be so sure that X-coordinate of center and point on X-Axis will always be same?
I mean if we are assuming them to be equal in derivation of a general equation then both must be same in all cases. But why? How can I prove that both center and X-Axis will have same X-coordinate?

A:

Simple geometry is a lot easier than algebra in this case. The $x$-axis is tangent to the circle, so the radius at the point of contact is perpendicular to the $x$-axis. Hence the centre and the point of contact have the same $x$-coordinate.

