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 \(\angle O_{1}O_{3}O_{2}\) is 90°, \(r_{3} =\) _____ cm.
[In the figure each circle touches the other two externally: \(C_{1}\) and \(C_{2}\) touch side by side, \(C_{3}\) sits above the point where they meet, and \(O_{3}\) is joined to \(O_{1}\) and to \(O_{2}\) with a right angle marked at \(O_{3}\).]
Answer: A
The radius is \(r_{3} = \frac{1}{2}(-3 + \sqrt{17})\) cm.
When two circles touch externally, the distance between their centres is the sum of their radii. So \(O_{1}O_{2} = 2 + 1 = 3\), \(O_{1}O_{3} = 2 + r_{3}\) and \(O_{2}O_{3} = 1 + r_{3}\). The right angle at \(O_{3}\) makes \(O_{1}O_{2}\) the hypotenuse of triangle \(O_{1}O_{3}O_{2}\):
\[ \begin{aligned} (2 + r)^{2} + (1 + r)^{2} &= 3^{2} \\ 2r^{2} + 6r + 5 &= 9 \\ r^{2} + 3r - 2 &= 0 \end{aligned} \]
\[ \begin{aligned} r &= \frac{-3 \pm \sqrt{9 + 8}}{2} \\ &= \frac{-3 \pm \sqrt{17}}{2} \end{aligned} \]
A radius must be positive, so \(r_{3} = (-3 + \sqrt{17})/2 \approx 0.56\) cm, which fits the small circle in the figure. \((3 + \sqrt{17})/2\) has the wrong sign on 3, \((-2 + \sqrt{17})/2\) has the wrong linear coefficient, and \((-3 + 2\sqrt{17})/2 \approx 2.6\) would make \(C_{3}\) larger than \(C_{1}\).
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