Discussion
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 \(\angle O_1O_3O_2\) is 90°, \(r_3\) = _____ cm.
(The figure shows the three circles each touching the other two externally, with \(O_3\) joined to \(O_1\) and to \(O_2\) and the right angle marked at \(O_3\).)
Answer: A
The radius is \(r_3 = \frac{1}{2}(-3 + \sqrt{17})\) cm.
Circles that touch externally have their centres separated by the sum of the radii, so the sides of triangle \(O_1O_3O_2\) are \(O_1O_3 = 2 + r_3\), \(O_2O_3 = 1 + r_3\) and \(O_1O_2 = 2 + 1 = 3\). The right angle is at \(O_3\), so by Pythagoras:
\[ \begin{aligned} (2 + r)^2 + (1 + r)^2 &= 3^2 \\ 2r^2 + 6r + 5 &= 9 \\ r^2 + 3r - 2 &= 0 \\ r &= \frac{-3 + \sqrt{17}}{2} \end{aligned} \]
taking the positive root (about 0.56 cm). The option with \(+3\) is the negative root with its sign dropped and is far too large; \(-2\) and \(2\sqrt{17}\) do not satisfy the quadratic.
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