As shown in the figure, circle \(C_1\) with center \(O_1\) and radius \(r_1\) touches the square \(VWXY\) at points \(P\) and \(Q\) while circle \(C_2\) with center \(O_2\) and radius \(r_2\) touches the square \(VWXY\) at points \(R\) and \(S\). The two circles touch each other at \(T\).
Given \(r_1 = 1\) cm and \(\overline{VY} = \overline{VW} = 4\) cm, \(r_2 =\) _____ cm.
Show Hint
Place the square in a coordinate system with W at the origin. Both circle centers then lie on the diagonal WY, at perpendicular distances r1 and r2 from the two nearby sides. Set the center to center distance equal to r1 plus r2.
Step 1: Notice both circle centers lie on the square's diagonal.
Circle $C_1$ is tangent to the two sides meeting at corner $W$, so its center $O_1$ is equally offset from both sides, which places $O_1$ exactly on the diagonal running from $W$ to $Y$. In the same way, circle $C_2$ is tangent to the two sides meeting at corner $Y$, so its center $O_2$ also lies on that same diagonal $WY$. This means $W$, $O_1$, $O_2$, $Y$ are all collinear, and we can work along this one line instead of using full coordinates.
Step 2: Find the length of the diagonal $WY$.
The square has side 4 cm, so by the Pythagorean theorem the diagonal is
\[ WY = \sqrt{4^2+4^2} = 4\sqrt{2} \text{ cm} \]
Step 3: Find how far $O_1$ is from corner $W$ along the diagonal.
$O_1$ sits a perpendicular distance $r_1$ from each of the two sides at $W$. For a point equally offset by distance $r_1$ from both sides of a right angle, the distance measured along the diagonal from the corner is $r_1\sqrt{2}$, the same 45 degree geometry as the diagonal of a small $r_1 \times r_1$ square. With $r_1 = 1$, that distance is $\sqrt{2}$ cm.
Step 4: Find how far $O_2$ is from corner $Y$ along the diagonal.
By identical reasoning at corner $Y$, $O_2$ sits a distance $r_2\sqrt{2}$ back from $Y$ along the diagonal.
Step 5: Express the gap between $O_1$ and $O_2$ along the diagonal.
The full diagonal $WY$ splits into three pieces: from $W$ to $O_1$, length $r_1\sqrt{2}$; from $O_1$ to $O_2$, the unknown center to center distance; and from $O_2$ to $Y$, length $r_2\sqrt{2}$. So
\[ O_1O_2 = 4\sqrt{2} - r_1\sqrt{2} - r_2\sqrt{2} = \sqrt{2}(4 - r_1 - r_2) \]
With $r_1=1$: $O_1O_2 = \sqrt{2}(3-r_2)$.
Step 6: Apply the external tangency condition and solve.
Since the circles touch each other, $O_1O_2 = r_1+r_2 = 1+r_2$. So
\[ \sqrt{2}(3-r_2) = 1+r_2 \]
\[ 3\sqrt{2}-1 = r_2(1+\sqrt{2}) \]
\[ r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}} = 7-4\sqrt{2} \text{ cm, after rationalizing} \]
Final Answer:
Working along the diagonal instead of full coordinates gives the same result.
\[ \boxed{r_2 = 7-4\sqrt{2} \text{ cm}} \]
