Consider a circle of radius R with center O.
Two other circles are internally tangent to this circle and intersect at points A and B.
Find the sum of the radii of the other two circles, given that ∠OAB = 90°.
R = short answer = AO_1 + AO_2
Let points D and E denote the common tangent points where the big circle touches the two small circles, and point C be the intersection point of the tangent lines at the points D and E.
First, there is a circle which passes through the points O, D, C, E, and A, because the line segment OC is seen from the points D, E, A under right angle.
Next, angle O_2AE = angle AEO_2 = say = a = angle ADO = angle DAO_1, thus angle OO_1A = angle OO_2A = 2a.
Second, angle OAO_2 = angle AOO_1 because
angle OAO_2 = angle OAE - a = is measured by the following arcs = ECDO/2 - a = EC/2 + CDO/2 - a = DC/2 + CDO/2 - a = DC/2 + pi/2 - a,
and similarly,
angle AOO_1 = measured by the arc DCEA/2 = DC/2 + CEA/2 + AO/2 - AO/2 = DC/2 + CEAO/2 - AO/2 = DC/2 + pi/2 - a.
Hence, triangles AOO_1 and AOO_2 are identical - they have a common side and 2 equal angles.
Thus, AO_1 = OO_2, so AO_1 + AO_2 = OO_2 + AO_2 = OO_2 + EO_2 = OE = R.
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