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.

Question

Consider a spherical globe rotating about an axis passing through its poles. There are three points P, Q, and R situated respectively on the equator, the north pole, and midway between the equator and the north pole in the northern hemisphere. Let P, Q, and R move with speeds $v_p$, $v_q$, and $v_r$, respectively.
Which one of the following options is CORRECT?

Answer 1

Here is a second way to reach the same value, using coordinates instead of plugging straight into the Pythagorean theorem.

Place $O_3$ at the origin, with the two radii to $O_1$ and $O_2$ running along the two coordinate axes, since the angle between them is $90^\circ$. So $O_1$ sits at $(-(r_1+r_3), 0)$, that is $(-(2+r_3), 0)$, and $O_2$ sits at $(0, -(r_2+r_3))$, that is $(0, -(1+r_3))$, using the tangency rule that touching circles have their centers separated by the sum of the two radii.
The distance from $O_1$ to $O_2$ must equal $r_1 + r_2 = 3$, because $C_1$ and $C_2$ also touch each other externally. Using the distance formula between $O_1$ and $O_2$: $\sqrt{(2+r_3)^2 + (1+r_3)^2} = 3$.
Square both sides: $(2+r_3)^2 + (1+r_3)^2 = 9$, which expands to $4 + 4r_3 + r_3^2 + 1 + 2r_3 + r_3^2 = 9$, so $2r_3^2 + 6r_3 - 4 = 0$, or $r_3^2 + 3r_3 - 2 = 0$.
Solve with the quadratic formula: $r_3 = \dfrac{-3 \pm \sqrt{17}}{2}$. A radius cannot be negative, and the figure shows $C_3$ as clearly the smallest circle, so take the root that gives a small positive value: $r_3 = \dfrac{-3+\sqrt{17}}{2} \approx 0.56$ cm.

Let's summarize:

Placing $O_3$ at the right angle and the other two centers along the axes turns the tangency conditions into a distance formula equation.
Squaring and simplifying gives the same quadratic $r_3^2 + 3r_3 - 2 = 0$ as the direct Pythagorean route.
Only the positive root, about 0.56 cm, matches a circle smaller than both $C_1$ and $C_2$, as shown in the figure.

So $r_3 = \dfrac{1}{2}\left(-3+\sqrt{17}\right)$ cm.

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

Answer 2

Here is a second way to reach the same value, using coordinates instead of plugging straight into the Pythagorean theorem.

Place $O_3$ at the origin, with the two radii to $O_1$ and $O_2$ running along the two coordinate axes, since the angle between them is $90^\circ$. So $O_1$ sits at $(-(r_1+r_3), 0)$, that is $(-(2+r_3), 0)$, and $O_2$ sits at $(0, -(r_2+r_3))$, that is $(0, -(1+r_3))$, using the tangency rule that touching circles have their centers separated by the sum of the two radii.
The distance from $O_1$ to $O_2$ must equal $r_1 + r_2 = 3$, because $C_1$ and $C_2$ also touch each other externally. Using the distance formula between $O_1$ and $O_2$: $\sqrt{(2+r_3)^2 + (1+r_3)^2} = 3$.
Square both sides: $(2+r_3)^2 + (1+r_3)^2 = 9$, which expands to $4 + 4r_3 + r_3^2 + 1 + 2r_3 + r_3^2 = 9$, so $2r_3^2 + 6r_3 - 4 = 0$, or $r_3^2 + 3r_3 - 2 = 0$.
Solve with the quadratic formula: $r_3 = \dfrac{-3 \pm \sqrt{17}}{2}$. A radius cannot be negative, and the figure shows $C_3$ as clearly the smallest circle, so take the root that gives a small positive value: $r_3 = \dfrac{-3+\sqrt{17}}{2} \approx 0.56$ cm.

Let's summarize:

Placing $O_3$ at the right angle and the other two centers along the axes turns the tangency conditions into a distance formula equation.
Squaring and simplifying gives the same quadratic $r_3^2 + 3r_3 - 2 = 0$ as the direct Pythagorean route.
Only the positive root, about 0.56 cm, matches a circle smaller than both $C_1$ and $C_2$, as shown in the figure.

So $r_3 = \dfrac{1}{2}\left(-3+\sqrt{17}\right)$ cm.

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

Show Hint

The Correct Option is A

Solution and Explanation

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