Radius and coordinates of orthogonal circle 
Let $C$ be a circle of radius $r$ centred on $(0,0)$. Let $a\in\Bbb R$ such that $0<a<r$. Find the center $(b,0)$ and the radius $s$ of the circle passing through $(a,0)$ and being orthogonal to $C$.

My approach:
Consider the radius vector of $C$: $(r\cos\theta, r\sin\theta)$. The radius vector of an orthogonal circle $D$ is $(-r\sin\theta, r\cos\theta)$. Since $D$ must pass through $(a,0)$, it has the radius vector $(b-a, 0)$ of length $|b-a|$. So the radius vector of $D$ is $(-|b-a|\sin\theta, |b-a|\cos\theta)$.
In my understanding the next crucial step is to find the angle $\theta$ of the radius vector of $C$. One train of thought I've considered is that we can drop the perpendicular vector at $(a, a\tan\theta)$, which passes through $(a,0)$ orthogonally. But I'm not sure what to do next. Would appreciate some hints. I'm a newbie to analytic geometry.
 A: 
If the two circles meet orthogonally at a point, the radius of one circle is tangent to the other, yielding a right-angled triangle. Then the Pythagorean theorem yields
$$r^2+(b-a)^2=b^2$$
$$r^2-2ab+a^2=0$$
$$b=\frac{r^2+a^2}{2a}$$
$$s=b-a=\frac{r^2-a^2}{2a}$$
A: Continuing with your approach, we know that the point $(b,0)$ must lie on the tangents to the first circle at the points of intersection of the two circles. We also know that $b\gt r\gt a$, otherwise the two circles don’t intersect. The equation of the tangent at $(r\cos\theta,r\sin\theta)$ is $x\cos\theta+y\sin\theta=r$, and the $x$-intercept of this line is $b=r\sec\theta$ (which you could also get by trigonometry). The distance from this point to the intersection point is equal to the radius of the second circle, $b-a$, from which we have $$(r\cos\theta-r\sec\theta)^2+r^2\sin^2\theta=(r\sec\theta-a)^2$$ which simplifies to $$2ar\sec\theta=r^2+a^2.$$ You could solve for $\theta$ at this point, but since $b=r\sec\theta$ you may as well substitute that into the equation and solve for $b$ directly, yielding $b={r^2+a^2\over2a}$ just as in another answer. (The method in that other answer is a much more direct way to get to this equation.)
