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.
Show 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.
This is a tangent circle problem that turns into a simple right triangle once you place the centers using the tangency rule.
Place $O_3$ at the origin with the right angle there. Since $C_1$ and $C_3$ touch externally, $O_3O_1 = r_1 + r_3 = 2 + r_3$. Since $C_2$ and $C_3$ touch externally, $O_3O_2 = r_2 + r_3 = 1 + r_3$. Put $O_1$ along the x-axis and $O_2$ along the y-axis so the $90^{\circ}$ angle at $O_3$ is automatic. Then $O_1$ sits at $(2+r_3, 0)$ and $O_2$ sits at $(0, 1+r_3)$.
Since $C_1$ and $C_2$ also touch externally, the straight-line distance $O_1O_2$ must equal $r_1 + r_2 = 3$. Using the distance formula:
$O_1O_2^2 = (2+r_3)^2 + (1+r_3)^2 = 9$
Expanding gives $2r_3^2 + 6r_3 - 4 = 0$, or $r_3^2 + 3r_3 - 2 = 0$. The quadratic formula gives $r_3 = \dfrac{-3 \pm \sqrt{17}}{2}$. The negative root makes $r_3$ come out negative, which is not possible for a radius, so only the plus root works.
This gives $r_3 = \dfrac{1}{2}(-3+\sqrt{17})$ cm, so option (A) is the answer.
