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 \(\angle O_1O_3O_2\) is \(90^\circ\), \(r_3 = \) cm.

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

Hint:

Since each pair of circles touches externally, the distance between any two centers equals the sum of their radii, then use this with the right angle at O3 and the Pythagorean theorem.

Answer 1

The Correct Option is A

Step 1: Turn circles touch each other into distances between centers.
When two circles touch each other from the outside, the distance between their centers equals the sum of their two radii.
So \( O_1O_3 = r_1 + r_3 \), \( O_2O_3 = r_2 + r_3 \), and \( O_1O_2 = r_1 + r_2 \).
Since \( r_1 = 2 \) and \( r_2 = 1 \), the distance \( O_1O_2 = 2 + 1 = 3 \) cm, a fixed number that does not depend on \( r_3 \).

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 a right triangle with the right angle at \( O_3 \).
The side opposite this right angle, \( O_1O_2 \), is the hypotenuse of that triangle, and \( O_1O_3 \) and \( O_2O_3 \) are the two legs.

Step 3: Apply the Pythagorean theorem.
\[ O_1O_2^2 = O_1O_3^2 + O_2O_3^2 \]
Substitute the distances from Step 1,
\[ 3^2 = (2 + r_3)^2 + (1 + r_3)^2 \]

Step 4: Expand both squared terms.
\[ (2+r_3)^2 = 4 + 4r_3 + r_3^2 \]
\[ (1+r_3)^2 = 1 + 2r_3 + r_3^2 \]
Adding these gives
\[ 9 = 5 + 6r_3 + 2r_3^2 \]

Step 5: Bring everything to one side and simplify.
\[ 2r_3^2 + 6r_3 + 5 - 9 = 0 \]
\[ 2r_3^2 + 6r_3 - 4 = 0 \]
Dividing every term by 2,
\[ r_3^2 + 3r_3 - 2 = 0 \]

Step 6: Solve the quadratic with the quadratic formula.
\[ r_3 = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2} = \frac{-3 \pm \sqrt{9+8}}{2} = \frac{-3 \pm \sqrt{17}}{2} \]
A radius cannot be negative, and \( \sqrt{17} \approx 4.12 \), so the minus root gives a negative number and must be thrown away.
\[ r_3 = \frac{-3 + \sqrt{17}}{2} \approx 0.56 \text{ cm} \]

Step 7: Why the other options are wrong.
Option (B) flips the sign in front of the 3, giving \( \frac{1}{2}(3+\sqrt{17}) \), which would only happen if the linear term in the quadratic came out negative instead of positive, a sign slip.
Option (C) uses a 2 in place of the 3, as if only one leg of the triangle had been squared correctly.
Option (D) doubles the square root term by mistake, which does not come from solving \( r_3^2 + 3r_3 - 2 = 0 \) correctly.

Final Answer:
\[ \boxed{r_3 = \frac{1}{2}\left(-3+\sqrt{17}\right) \text{ cm}} \]