Circles \(C_1\), \(C_2\), and \(C_3\), with centers \(O_1\), \(O_2\), and \(O_3\), and radii \(r_1\), \(r_2\), and \(r_3\), respectively, touch each other as shown in the following figure. Given \(r_1 = 2\) cm, \(r_2 = 1\) cm and the angle \(\angle O_1 O_3 O_2\) is 90 degrees, \(r_3 =\) ______ cm.
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Since the circles are mutually tangent, each side of triangle O1O3O2 is a sum of two radii; the given right angle then lets you apply Pythagoras directly.
Step 1: Express the side lengths of triangle O1O3O2 using the tangency condition. When two circles touch each other externally, the distance between their centers equals the sum of their radii. Since \(C_1\) and \(C_3\) touch, \(O_1O_3 = r_1 + r_3\). Since \(C_2\) and \(C_3\) touch, \(O_2O_3 = r_2 + r_3\). Since \(C_1\) and \(C_2\) touch, \(O_1O_2 = r_1 + r_2\).
Step 2: Substitute the known radii. With \(r_1 = 2\) and \(r_2 = 1\): \(O_1O_3 = 2 + r_3\), \(O_2O_3 = 1 + r_3\), and \(O_1O_2 = 2 + 1 = 3\).
Step 3: Use the right angle at O3. We are told \(\angle O_1O_3O_2 = 90^\circ\), so triangle \(O_1O_3O_2\) is right-angled at \(O_3\), meaning \(O_1O_2\) is the hypotenuse. By the Pythagorean theorem: \[(O_1O_3)^2 + (O_2O_3)^2 = (O_1O_2)^2\]
Step 5: Simplify into a standard quadratic. \[2r_3^2 + 6r_3 - 4 = 0\] \[r_3^2 + 3r_3 - 2 = 0\]
Step 6: Solve using the quadratic formula. \[r_3 = \frac{-3 \pm \sqrt{9+8}}{2} = \frac{-3 \pm \sqrt{17}}{2}\]
Step 7: Reject the negative root. A radius must be positive, and \(\sqrt{17} \approx 4.12\), so \(\frac{-3-\sqrt{17}}{2}\) is negative and physically meaningless. The only valid solution is \[r_3 = \frac{-3+\sqrt{17}}{2} = \frac{1}{2}\left(-3+\sqrt{17}\right)\] which is option (A).
