This may not be much help to you, but have you considered changing coordinates so that the centre of the circle is at 0,0. Then at a point (a,b) on the circle you have aa+bb=rr

where r is the radius of the circle and the equation of the tangent there is ax+by-rr=0.

You will probably also want to put a=r cos theta and b = r sin theta as you imagine yourself traveling round the circle and theta goes from 0 to two*pi radians, and then maybe use the half angle formulas for sin and cos in terms of t = tan(theta/2) to get rid of the trig functions.

Then you can substitute those values into the (new) equation of the ellipse to find the intersections of the circle tangent at (a,b) with the ellipse. That should get you a quadratic equation in t, and the common tangents will be the points where the quadratic has equal real roots, so you can use the usual formula for solving quadratics and find where the bit you take the square root of is zero.

You could probably do things the other way around, and that might be easier but this way you don't have to know the general rule for the equation of a tangent to an ellipse.