The radius of the circle which just touches the outer circle

Question

Suppose two concentric circles with radius $a$ and $b\ (>a)$ and origin as their center.

I wanted to put another circle whose center lies in a line $x=a$ (that is red line) in such a way that it just touches the outer circle at a single point. Something like this

As an equation solving a problem we can solve $$(x-a)^2+(y-r)^2=r^2$$ $$x^2+y^2=b^2$$ find the quadratic in $x$ and make the roots equal to each other. But that's too long a calculation, I wanted to know if I can solve it from the geometry of the figure. Or some much simpler method. Mainly I'm interested in $r$. Can any help me with this?

Answer 1

So the following is the idea commented by @Kaind, which uses geometry to solve the problem. By the way, By geometry, I don't mean taking scale and pencil and then go about taking measurements.

In triangle $\Delta ACD$ $$(AC)^2=a^2+r^2$$ Also $$AC+CB=AB\Rightarrow \sqrt{a^2+r^2}+r=b$$ $$(a^2+r^2)=(b-r)^2$$ $$\boxed{r=\frac{b^2-a^2}{2b}}$$

Answer 2

If the red circle is to touch the blue one, then both their centers and the tangency point lie on a common line. So, draw a line through the center of the blue circle. Its intersection with the red line defines a center of the red circle, and its intersection with the blue circle defines a tangency point. Their distance is a radius you need.