Two Circles and a Triangle

Draw a straight line BQ from B to a point Q on AC. Now, CQ and CA are tangents to the bigger and smaller circle separately.We know that tangents drawn from an external point to a particular circle are equal.Therefore,AQ = BQ and BQ = CQ. So,AQ = CQ. Now, ABC is a triangle in which angle QAB + angle QBA + angle QBC + angle QCB = 180 degree..Since QA = QB and QB = QC,angle QZB = angle QBA and angle QBC = angle QCB.Therefore,2(angle QBA + angle QBC) = 180 degree.Therefore,angle ABC = 90 degree.

Draw a line that goes through ${B}$ and the centers of both circles. Let that line intersect the smaller circle at ${P}$ and the bigger one at ${Q}$ . Also, let line AC extend to the left to point ${x}$ and to the right to point ${x}$ . Now angles in a semicircle equal 90 degrees so: $\angle{PAB} = \angle{BCQ} = 90$ $Now \ let \angle{PAX} = a \ and \angle{QCY} = b$ . $using \ alternate \ segment \ theorem$ $\angle{QBC}\ = b\ and \angle{PBA}\ = a\Rightarrow\angle{ABC}\ = 180 - (a+b)\ (1)$ . By using angle on a straight line = 180 degrees and some angle - chasing gives: $\angle{ABC}\ = 180 - (90-a) - (90-b) = a+b\ (2)$ Now equating (1) and (2): $(a+b) = 180 - (a+b)\Rightarrow\ (a+b) = 90. \therefore\angle{ABC} = 90$

Let o, O be the center of small, large circles, respectively. Draw in the lines Ao, oO, OC and AoOC make a quadrilateral. We know the sum of the angles of a quadrilateral is 360° Clearly, ∠oAC and ∠OCA are both 90°, and that ∠oBA +∠OBC = 90°, since ∆ oAB and ∆OBC are isosceles. ∴ ∠ABC = 90°

since they don't specify the radii of the circles, we can assume that this is true for all circles. So let the two circles have the same radii. The centres of the circles and A and B form a rectangle, so an angle ABC must be equal to 90 degrees.