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.
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Since the circles are mutually tangent, each side of triangle O1O3O2 is a sum of two radii; the given right angle then lets you apply Pythagoras directly.
A quicker route is to place the triangle on coordinate axes using the right angle itself as the origin. Since $\angle O_1O_3O_2 = 90^\circ$, put $O_3$ at the origin $(0,0)$ with $O_1O_3$ running along the x-axis and $O_2O_3$ running along the y-axis. Because the circles touch, $O_1O_3 = r_1 + r_3 = 2 + r_3$, so place $O_1$ at $(2+r_3, 0)$. Similarly $O_2O_3 = r_2 + r_3 = 1 + r_3$, so place $O_2$ at $(0, 1+r_3)$.
Now compute the straight-line distance between $O_1$ and $O_2$ using the distance formula, which must equal $O_1O_2 = r_1 + r_2 = 3$ since $C_1$ and $C_2$ also touch each other:
\[ \sqrt{(2+r_3)^2 + (1+r_3)^2} = 3 \]
Squaring both sides removes the root:
\[ (2+r_3)^2 + (1+r_3)^2 = 9 \]
Expanding gives $4 + 4r_3 + r_3^2 + 1 + 2r_3 + r_3^2 = 9$, which collects to $2r_3^2 + 6r_3 - 4 = 0$, or after dividing by 2, $r_3^2 + 3r_3 - 2 = 0$. Applying the quadratic formula, $r_3 = \dfrac{-3 \pm \sqrt{17}}{2}$.
A circle cannot have a negative radius, and $\sqrt{17}\approx4.12$, so the minus branch gives a negative number and is discarded, leaving the plus branch as the physical answer.
