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 \(90^{\circ}\), \(r_3 =\) _____ cm.
Show Hint
Use the fact that touching circles have their centers separated by the sum of their radii, then apply the Pythagorean theorem to the right angle at \(O_3\).
Step 1: Use the tangency condition to write the three side lengths.
Since \(C_1\) and \(C_2\) touch each other from outside, the distance between their centers equals the sum of their radii: \(O_1O_2 = r_1 + r_2\). The same rule applies to the other two pairs, since all three circles touch each other externally:
\[ O_1O_2 = r_1 + r_2, \quad O_1O_3 = r_1 + r_3, \quad O_2O_3 = r_2 + r_3 \]
Step 2: Apply the right angle at \(O_3\).
Triangle \(O_1O_2O_3\) has a right angle at \(O_3\), so by the Pythagorean theorem, the side opposite the right angle, \(O_1O_2\), satisfies:
\[ O_1O_2^2 = O_1O_3^2 + O_2O_3^2 \]
Step 4: Solve the quadratic equation.
\[ 2r_3^2 + 6r_3 - 4 = 0 \]
\[ r_3^2 + 3r_3 - 2 = 0 \]
Using the quadratic formula:
\[ r_3 = \frac{-3 \pm \sqrt{9+8}}{2} = \frac{-3 \pm \sqrt{17}}{2} \]
A radius must be positive, and the figure shows \(C_3\) as clearly the smallest of the three circles, so \(r_3\) must be smaller than both \(r_1=2\) and \(r_2=1\). Only one root fits both conditions:
\[ r_3 = \frac{-3+\sqrt{17}}{2} \approx 0.56 \text{ cm} \]
The other root, \(\frac{-3-\sqrt{17}}{2}\), is negative and cannot be a radius. Option (B), \(\frac{1}{2}(3+\sqrt{17}) \approx 3.56\) cm, is bigger than \(r_1\) itself, which does not match a circle drawn as the smallest of the three. Options (C) and (D) do not solve the quadratic equation above at all.
Final Answer:
\(r_3 = \dfrac{1}{2}\left(-3+\sqrt{17}\right)\) cm.
\[ \boxed{r_3 = \frac{1}{2}\left(-3+\sqrt{17}\right)} \]
