Finding the circles passing through two points and touching a circle Given two points and a circle, construct a/the circle through the two points and
touching the given circle.
I came across this problem in History of Numerical Analysis by H. Goldstein. I
spent some time on this. I have a method of constructing it using radical axis.
I am wondering if there is a more elementary construction.
 A: Since your proposed solution in the comments doesn't always work, let me give a somewhat more detailed description of the construction than the one outlined by user8268 (I'm assuming the given points are outside the given circle):

when unsure always use circle inversion :) (here wrt. a circle with the center in one of the 2 point - after inversion the point goes to infinity, so you now need to find a line passing though a point (the image of the other point) and touching a circle (the image of the original circle). Then apply the inversion again, and the image of the line is the circle you're looking for.




*

*Given: Two points $\color{blue}{A},\color{blue}{B}$ and a circle $\color{blue}{c}$ with center $\color{blue}{C}$.

*Wanted: The two points $\color{red}{P,Q}$ on $\color{blue}{c}$ such that the circles through $\color{blue}{A},\color{blue}{B},\color{red}{P}$ and $\color{blue}{A},\color{blue}{B},\color{red}{Q}$ are tangent to $\color{blue}{c}$ (drawing a circle through three points is trivial).


Construction:
The idea is outlined in user8268's comment: Draw a circle $\color{lime}{e}$ around $\color{blue}{A}$ intersecting $\color{blue}{c}$ orthogonally. The point is that $\color{blue}{c}$ is invariant under circle reflection at $\color{lime}{e}$. The point $\color{blue}{A}$ is sent to infinity and the sought circles will therefore become tangents to $\color{blue}{c}$. Since the circles pass through $\color{blue}{B}$, the tangents must pass through the reflection $\color{blue}{B'}$ of $\color{blue}{B}$ at $\color{lime}{e}$. Thus here's the construction in brief:


*

*Construct the circle $\color{lime}{e}$ around $\color{blue}{A}$ intersecting $\color{blue}{c}$ orthogonally.

*Reflect $\color{blue}{B}$ at $\color{lime}{e}$ to get $\color{blue}{B'}$.

*Find the tangents from $\color{blue}{B'}$ to $\color{lime}{e}$.

*and 5: reflect the tangents at $\color{lime}{e}$ to find the circles.


So here we go in more detail:


*

*Draw the circle $\color{lime}{e}$ with center $\color{blue}{A}$ intersecting the circle $\color{blue}{c}$ orthogonally: First draw the circle with diameter $\color{blue}{AC}$ (dashed circle below) and intersect it with $\color{blue}{c}$, then draw the circle $\color{lime}{e}$ with center $\color{blue}{A}$ through those intersection points. The radii of $\color{blue}{c}$ and $\color{lime}{e}$ will meet perpendicularly at the points of intersection by Thales, as $\color{blue}{AC}$ is the diameter of the dashed circle. 

*Reflect $\color{blue}{B}$ at the circle $\color{lime}{e}$ to obtain $\color{blue}{B'}$. Note that $\color{blue}{B'}$ must lie outside $\color{blue}{c}$ as $\color{blue}{B}$ lies outside $\color{blue}{c}$. Recall the construction if $\color{blue}{B}$ lies outside $\color{lime}{e}$: the point of intersection of the segment $\color{blue}{AB}$ with the segment through the points of intersection of  $\color{lime}{e}$ with the circle through $\color{blue}{A}$ and $\color{blue}{B}$ around their midpoint:
Exercise: Give the construction if $\color{blue}{B}$ lies inside $\color{lime}{e}$.

*Find the points of contact $P'$ and $Q'$ of the tangents from $\color{blue}{B'}$ to $\color{blue}{c}$: Draw the circle $\color{orange}{d}$ with diameter $\color{blue}{B'C}$ and call $P'$ and $Q'$ the points of intersection of $\color{orange}{d}$ with $\color{blue}{c}$ (see step 1.): 

*Since reflection at $\color{lime}{e}$ swaps $A$ and infinity and leaves the circle $\color{blue}{c}$ invariant, the tangents from $\color{blue}{B'}$ to $\color{blue}{c}$ will become circles tangent to $\color{blue}{c}$ upon reflection  at $\color{lime}{e}$. Therefore the points $\color{red}{P}$ and $\color{red}{Q}$ are the reflections of $P'$ and $Q'$ at $\color{lime}{e}$. Since $\color{blue}{c}$ is orthogonal to $\color{lime}{e}$ they are simply the second points of intersection of $\color{blue}{c}$ with the lines $\color{blue}{A}P'$ and $\color{blue}{A}Q'$ (since those lines and the circle $\color{blue}{c}$ are invariant under reflection at $\color{lime}{e}$):

*Finally, draw the circles through $\color{blue}{A},\color{blue}{B},\color{red}{P}$ and $\color{blue}{A},\color{blue}{B},\color{red}{Q}$:

voilà!

Exercise: Treat the case in which both points $\color{blue}{A}$ and $\color{blue}{B}$ are inside the circle $\color{blue}{c}$.
Hint: Reflect at the circle with center $\color{blue}{B}$ through $\color{blue}{A}$.

Added: In retrospect I like the second solution using reflection at the circle around $\color{blue}{B}$ through $\color{blue}{A}$ better than the one I gave, because it works in all cases that have a solution—either $\color{blue}{A}$ and $\color{blue}{B}$ both inside or both outside of the circle $\color{blue}{c}$. At some point I might add a detailed explanation of that as a second answer.
A: Here are some pictures illustrating the argument I alluded to at the end of my first answer. It has the virtue of working without distinguishing the cases whether the two points $\color{blue}{A}$ and $\color{blue}{B}$ are inside or outside the given circle $\color{blue}{c}$.


*

*Given: Two points $\color{blue}{A},\color{blue}{B}$ and a circle $\color{blue}{c}$ with center $\color{blue}{C}$.

*Wanted: The two points $\color{red}{P,Q}$ on $\color{blue}{c}$ such that the circles through $\color{blue}{A},\color{blue}{B},\color{red}{P}$ and $\color{blue}{A},\color{blue}{B},\color{red}{Q}$ are tangent to $\color{blue}{c}$ (drawing a circle through three points is trivial).


Before delving into the solution let's get rid of some degenerate cases:
Warm-up Exercise: Treat all the cases in which $\color{blue}{A} = \color{blue}{B}$ or one of $\color{blue}{A}$ or $\color{blue}{B}$ lies on $\color{blue}{c}$.
So, from this point on, we assume $\color{blue}{A} \neq \color{blue}{B}$ and that both lie either inside or outside the circle $\color{blue}{c}$.

The idea is the same as in the other answer (i.e. user8268's solution). Reflecting the configuration at the circle $\color{green}{d}$ with center  $\color{blue}{B}$ through $\color{blue}{A}$ fixes $\color{blue}{A}$ and sends $\color{blue}{B}$ to infinity. It transforms the circle $\color{blue}{c}$ into a circle $\color{blue}{c'}$ and transforms the circles we're looking for into tangents from $\color{blue}{A}$ to $\color{blue}{c'}$ because $\color{blue}{B}$ is sent to infinity and circle reflection preserves angles. Finding the tangents from the point $\color{blue}{A}$ to the circle $\color{blue}{c'}$ is easy and we need only reflect the points of tangency $P',Q'$ back to $\color{blue}{c}$ to find $\color{red}{P}$ and $\color{red}{Q}$. Drawing the circles through $\color{blue}{A}, \color{blue}{B}, \color{red}{P}$ and $\color{blue}{A}, \color{blue}{B}, \color{red}{Q}$ is again straightforward.
So here's the solution in somewhat more detail: 


*

*Reflect the circle $\color{blue}{c}$ at the circle $\color{green}{d}$ through $\color{blue}{A}$: 
To find $\color{blue}{c'}$, draw the line through $\color{blue}{B}$ and $\color{blue}{C}$ (if $\color{blue}{B} = \color{blue}{C}$ draw an arbitrary line through $\color{blue}{B}$) and reflect its points of intersection with $\color{blue}{c}$ at $\color{green}{d}$. Then the circle $\color{blue}{c'}$ is the circle with diameter those two reflected points. (See also point 2. in my other answer for more details.)Exercise: Prove that $\color{blue}{A}$ is always outside the circle $\color{blue}{c'}$.

*Find the tangents from $\color{blue}{A}$ to $\color{blue}{c'}$:  Call the points of tangency $P'$ and $Q'$.

*Reflect the points $P'$ and $Q'$ at $\color{green}{d}$ to find $\color{red}{P}$ and $\color{red}{Q}$: 

*Draw the circles through $\color{blue}{A}, \color{blue}{B}, \color{red}{P}$ and $\color{blue}{A}, \color{blue}{B}, \color{red}{Q}$:  done.
Remark: Note that the red circles pass through the second points of intersection of $\color{green}{d}$ with the tangents from $\color{blue}{A}$ to $\color{blue}{c'}$. This means that the construction admits a slightly simpler variant (omitting steps 3 and 4) if one doesn't care about the points $\color{red}{P}$ and $\color{red}{Q}$. I chose to explain it the way I did, as the more efficient method in this remark doesn't exhibit as clearly why it works and in order to see that, one needs to consider the points $\color{red}{P}$ and $\color{red}{Q}$ anyway.

As a final picture, the case that both $\color{blue}{A}$ and $\color{blue}{B}$ lie outside of $\color{blue}{c}$: 
