Given the angle $∠pOq$ and point $M$, which is not on the ray $p$. Construct a circle with its center on line $q$ that passes through point $M$ and touches line $p$.
If I mark the point of tangency on line p as D, and the center of the circle on line q as S. Then, I have triangle SMD which is isosceles, but I'm not sure how to proceed further.
I am sure the simpler solution exists, but for starters...
Let $p'$ be a reflection of $p$ with respect to $q$. The circle $C$ you are looking for is tangent to both $p$ and $p'$. This is obviously a variation on the Apollonius problem theme.
Pick any circle of an arbitrary radius, centered at $M$. Make an inversion. It will map both $p$ and $p'$ to some circles, and $C$ to a straight line. That means, you need to find a straight line touching two circles, and take the inversion back.
It can be done this way: Lets say that the center is C on $q$. Draw the line from C normal to the line $q$. Lets call the intersection N. Now you have isosceles triangle $MCN$. It should be easy to determine the angle at the point M in triangle $MCN$ in terms of $\angle{pOq}$. Denote that angle as $\alpha$. Now use the rotation $R_{(M,\alpha)}$. The point N will be mapped into the point C. Line $p$ will be mapped to line $p'$, meaning the center C will be intersection of $q$ and $p'$. All you need to do is to draw $R_{(M,\alpha)}(p)=p'$. The rest should be easy.
