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 degrees, \(r_3 =\) _____ 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:

Use tangent circle distances (sum of radii) with the Pythagorean theorem on the right angle at O3, then solve the resulting quadratic.

Answer 1

The Correct Option is A

Step 1: Place O3 at the right angle corner.
Since the angle at $O_3$ between the lines to $O_1$ and $O_2$ is $90^{\circ}$, it helps to put $O_3$ at the origin of a coordinate grid, with one arm along the x axis and the other along the y axis.
Let $O_3 = (0,0)$, $O_1 = (2+r_3, 0)$ since $O_1O_3 = r_1 + r_3 = 2 + r_3$, and $O_2 = (0, 1+r_3)$ since $O_2O_3 = r_2 + r_3 = 1 + r_3$.

Step 2: Write the distance between O1 and O2.
$C_1$ and $C_2$ also touch each other directly, so $O_1O_2$ must equal $r_1 + r_2 = 3$. Using the distance formula between the two points placed above:
\[ O_1O_2 = \sqrt{(2+r_3)^2 + (1+r_3)^2} = 3 \]

Step 3: Square both sides and solve.
\[ (2+r_3)^2 + (1+r_3)^2 = 9 \]
\[ 4 + 4r_3 + r_3^2 + 1 + 2r_3 + r_3^2 = 9 \]
\[ 2r_3^2 + 6r_3 - 4 = 0 \]
\[ r_3^2 + 3r_3 - 2 = 0 \]
\[ r_3 = \frac{-3 \pm \sqrt{17}}{2} \]

Step 4: Keep the physically valid root.
A circle radius must be a positive number. Since $\sqrt{17} \approx 4.12$, the root with the minus sign gives a negative number, which cannot be a radius. Only the positive branch works:
\[ r_3 = \frac{-3+\sqrt{17}}{2} \approx 0.56 \text{ cm} \]
This matches option (A). \[ \boxed{r_3 = \frac{-3+\sqrt{17}}{2} \text{ cm}} \]