Question

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.

• A. \( \dfrac{1}{2}(-3 + \sqrt{17}) \)
• B. \( \dfrac{1}{2}(3 + \sqrt{17}) \)
• C. \( \dfrac{1}{2}(-2 + \sqrt{17}) \)
• D. \( \dfrac{1}{2}(-3 + 2\sqrt{17}) \)

Hint:

Use external tangency to write each side of triangle O1O2O3 in terms of the radii, then apply the Pythagoras rule at the right angle.

Answer 1

The Correct Option is A

Step 1: Write the triangle sides using tangency.
When two circles touch each other from outside, the distance between their centers equals the sum of their radii.
So \( O_1O_2 = r_1 + r_2 = 2 + 1 = 3 \) cm.
\( O_1O_3 = r_1 + r_3 = 2 + r_3 \) cm and \( O_2O_3 = r_2 + r_3 = 1 + r_3 \) cm.

Step 2: Apply the right angle at O3.
The angle \( \angle O_1O_3O_2 = 90^{\circ} \) means triangle \( O_1O_3O_2 \) is right angled at \( O_3 \), with \( O_1O_2 \) as the hypotenuse.
By the Pythagoras rule: \( O_1O_2^2 = O_1O_3^2 + O_2O_3^2 \).

Step 3: Substitute and expand.
\( 3^2 = (2+r_3)^2 + (1+r_3)^2 \)
\( 9 = 4 + 4r_3 + r_3^2 + 1 + 2r_3 + r_3^2 \)
\( 9 = 5 + 6r_3 + 2r_3^2 \)
\( 2r_3^2 + 6r_3 - 4 = 0 \), which gives \( r_3^2 + 3r_3 - 2 = 0 \).

Step 4: Solve the quadratic.
\[ r_3 = \frac{-3 \pm \sqrt{9+8}}{2} = \frac{-3 \pm \sqrt{17}}{2} \] Since a radius must be positive, take the \( + \) root.

Final Answer:
\( r_3 = \dfrac{1}{2}(-3+\sqrt{17}) \) cm, so option (A) is correct. \[ \boxed{r_3 = \tfrac{1}{2}(-3+\sqrt{17})} \]