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.
Show Hint
Use tangent circle distances (sum of radii) with the Pythagorean theorem on the right angle at O3, then solve the resulting quadratic.
Step 1: Write the tangency conditions as distances.
When two circles touch each other from outside, the distance between their centers equals the sum of their radii. Applying this to each touching pair:
\[ O_1O_2 = r_1 + r_2 = 2 + 1 = 3 \]
\[ O_1O_3 = r_1 + r_3 = 2 + r_3 \]
\[ O_2O_3 = r_2 + r_3 = 1 + r_3 \]
Step 2: Use the right angle at \(O_3\).
We are told \(\angle O_1O_3O_2 = 90^{\circ}\), so triangle \(O_1O_3O_2\) is right angled at \(O_3\). By the Pythagorean theorem, the square of the side opposite the right angle equals the sum of the squares of the other two sides:
\[ O_1O_3^2 + O_2O_3^2 = O_1O_2^2 \]
\[ (2+r_3)^2 + (1+r_3)^2 = 3^2 \]
Step 4: Solve the quadratic.
Using the quadratic formula with \(a=1\), \(b=3\), \(c=-2\):
\[ r_3 = \frac{-3 \pm \sqrt{9+8}}{2} = \frac{-3 \pm \sqrt{17}}{2} \]
A radius cannot be negative, and \(\sqrt{17}\) is close to 4.12, so only the plus sign gives a positive value near 0.56 cm. So we keep:
\[ r_3 = \frac{-3+\sqrt{17}}{2} \]
Final Answer:
\(r_3 = \dfrac{1}{2}\left(-3+\sqrt{17}\right)\) cm, which is option (A). Option (B) flips the sign in front of 3. Option (C) uses -2 instead of -3 inside the bracket, a slip in the constant term. Option (D) doubles the surd term by mistake.
\[ \boxed{r_3 = \tfrac{1}{2}\left(-3+\sqrt{17}\right) \text{ cm}} \]
