Since the circle's radii form right angles with line joining the external point to the points of tangency at the points of tangency, the quadrilateral formed by joning the centre to one tangent point to the external point to the second tangency point and so back to the centre is a kite in which the opposite equal angles are both 90 degrees. Thus it is a cyclic quadralateral and furthermore the line from the original circle's centre to the external point is a diameter of this kite's circumcircle. Thus the two points of tangency are the intersections of the original circle and the circle centred at the midpoint of the line joining the original circle's centre to the external point, and with a radius of half its length.
> Can anyone tell me (if it exists) the formula to find a tangent point on > a Circle at (x,y) with radius R from any known point outside the Circle? > I've tried to accomplish this using the method outlined in this newsgroup > posting "Coordinate (I think) vs Polar Angles". It is best if you start > with the "Coordinate (I think) vs Polar Angles w/ correct graphic" posted by > Charlie. > > All Help is Greatly Appreciated, > email@example.com