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 $\color{blue}{c}$: