Constructing a circle that internally tangents a circle $\gamma$ and passes through two internal points. The full details of this problem is given as follows

Construct a circle $\gamma$ with center $O_\gamma$ ,  and 
    place two points $A$ and $B$ inside $\gamma$. That does not lie on the edge of the circle. Explain the construction of a point $C$, such that the circle $ABC =\beta$, is internally tangential to $\gamma$.

Now $ABC$ means a circle that passes through the points $A$,$B$ and $C$. I have made a drawing, but I am unable to mathematicaly construct the point $C$. 
I already know that for most pairs $A$,$B$ there are two possible choices for $C$. Eg $C_1$ and $C_2$. See the following figure

Can anyone show me or help me in finding the placement of $C$, given $A$ and $B$?
The figure is only but a sketch, but I know that the centre of the circle obviously has to lie on the perpendicular bisector of A and B, after that I am clueless.
 A: I gave an answer here, which applies to the general case of finding a circle tangent to three circles, which may have up to 8 solutions. This particular case is a degenerate one where two of the circles are points, and has only 2 solutions. It can be solved by the last step of the general solution I gave, which is to invert at one of the two points, and the desired circle will become a line that must be tangent to a circle and pass through a point, which is easy to construct. Inversion can be easily constructed using compass and straightedge also, and the original desired solutions can likewise be obtained by undoing the inversion.
A: I will give you the general solution of the apollonius PPC which works fine for this case:
draw any circle $\Gamma$ that passes through $A$ and $B$ and meets $\gamma$ in two points $M$ and $N$.
Let $H = MN \cap AB$ draw the tangents from $H$ to $\gamma$ and let $T$ and $S$ be the contact points of these tangent lines.
The circles you want are the circles $\odot (TAB)$ and $\odot (SAB)$.
The construction works because $H$ is by construction the radical center of $\gamma$, $\Gamma$ and the solutions and therefore lies in the common tangent line of the solutions and $\gamma$.
