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: Set up coordinates with the right angle at the origin.
Since $\angle O_1O_3O_2 = 90^\circ$, place $O_3$ at the origin with the two legs of the right angle running along the x-axis and y-axis.
Put $O_1$ on the x-axis at $(2+r_3, 0)$, because $O_1O_3 = r_1 + r_3 = 2 + r_3$ (circles $C_1$ and $C_3$ touch externally).
Put $O_2$ on the y-axis at $(0, 1+r_3)$, because $O_2O_3 = r_2 + r_3 = 1 + r_3$ (circles $C_2$ and $C_3$ touch externally).

Step 2: Write the distance between $O_1$ and $O_2$ using coordinates.
The distance formula between $(2+r_3, 0)$ and $(0, 1+r_3)$ gives
\[ O_1O_2 = \sqrt{(2+r_3)^2 + (1+r_3)^2} \]
This is really just the Pythagorean theorem again, but built directly from coordinates instead of a labeled triangle.

Step 3: Fix this distance using the fact that $C_1$ and $C_2$ also touch.
Circles $C_1$ and $C_2$ touch each other externally too, since they meet at a single point in the figure, so $O_1O_2 = r_1 + r_2 = 2 + 1 = 3$.
Setting the coordinate distance equal to 3,
\[ \sqrt{(2+r_3)^2 + (1+r_3)^2} = 3 \]

Step 4: Square both sides to remove the square root.
\[ (2+r_3)^2 + (1+r_3)^2 = 9 \]
\[ (4 + 4r_3 + r_3^2) + (1 + 2r_3 + r_3^2) = 9 \]
\[ 2r_3^2 + 6r_3 + 5 = 9 \]

Step 5: Rearrange into standard quadratic form.
\[ 2r_3^2 + 6r_3 - 4 = 0 \]
\[ r_3^2 + 3r_3 - 2 = 0 \]

Step 6: Solve for $r_3$ and pick the physical root.
Completing the square, $\left(r_3 + \frac{3}{2}\right)^2 = 2 + \frac{9}{4} = \frac{17}{4}$.
Taking the square root, $r_3 + \frac{3}{2} = \pm \frac{\sqrt{17}}{2}$, so $r_3 = \frac{-3 \pm \sqrt{17}}{2}$.
A circle's radius must be a positive length, and the minus option here works out negative, near $-3.56$, so only the plus option survives,
\[ r_3 = \frac{-3+\sqrt{17}}{2} \approx 0.56 \text{ cm} \]

Final Answer:
Coordinate geometry gives the same result, $r_3 = \frac{1}{2}(-3+\sqrt{17})$ cm, option (A).
\[ \boxed{r_3 = \frac{1}{2}\left(-3+\sqrt{17}\right)} \]