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.
Show Hint
The right angle at \(O_3\) makes \(O_1O_2\) the hypotenuse of triangle \(O_1O_3O_2\); apply the Pythagorean theorem with \(O_1O_2=3\), \(O_1O_3=2+r_3\), \(O_2O_3=1+r_3\).
Step 1: Understanding the Concept:
Place the right angle at $O_3$ on a coordinate grid instead of applying the Pythagorean theorem directly by name. Put $O_3$ at the origin, with the two legs of the right angle, $O_3O_1$ and $O_3O_2$, running along the x-axis and y-axis.
Step 2: Key Formula or Approach:
Since the circles touch each other externally, the distance between any two centers is the sum of the corresponding radii: $O_1O_2 = r_1+r_2$, $O_3O_1 = r_1+r_3$, $O_3O_2 = r_2+r_3$. With $O_3$ at the origin and the right angle sitting exactly at $O_3$, place $O_1$ at $(r_1+r_3, 0)$ along the x-axis and $O_2$ at $(0, r_2+r_3)$ along the y-axis.
Step 3: Detailed Explanation:
With $r_1 = 2$ and $r_2 = 1$, the coordinates become $O_1 = (2+r_3, 0)$ and $O_2 = (0, 1+r_3)$. The distance formula between these two points gives the length $O_1O_2$ directly:
$O_1O_2 = \sqrt{(2+r_3)^2 + (1+r_3)^2}$
But we also know $O_1O_2 = r_1+r_2 = 3$ from the tangency of circles $C_1$ and $C_2$ with each other. Setting the two expressions equal and squaring both sides:
$9 = (2+r_3)^2 + (1+r_3)^2 = 4+4r_3+r_3^2+1+2r_3+r_3^2 = 2r_3^2+6r_3+5$
$2r_3^2+6r_3-4 = 0$, which simplifies to $r_3^2+3r_3-2=0$.
Solving this quadratic: $r_3 = \dfrac{-3\pm\sqrt{9+8}}{2} = \dfrac{-3\pm\sqrt{17}}{2}$. A radius must be a positive length, so only the "+" root survives: $r_3 = \dfrac{-3+\sqrt{17}}{2}$.
Step 4: Final Answer:
Placing the right angle at the origin and using the distance formula gives the same result as the direct Pythagorean approach. \[ \boxed{r_3 = \frac{1}{2}\left(-3+\sqrt{17}\right) \text{ cm}} \]
