Question

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.

• A. \(4 - 3\sqrt{2}\)
• B. \(1 + 2\sqrt{2}\)
• C. \(7 - 4\sqrt{2}\)
• D. \(5 + 3\sqrt{2}\)

Hint:

Both circle centers lie on the diagonal joining the two corners the circles hug. Use the tangency condition that the distance between centers equals the sum of the radii.

Answer 1

The Correct Option is C

Step 1: Notice both centers lie on the same diagonal.
With \(W = (0,0)\) and \(Y = (4,4)\), the diagonal \(WY\) is the line $y = x$. Since \(O_1 = (r_1, r_1)\) and \(O_2 = (4-r_2, 4-r_2)\) both satisfy $y = x$, both circle centers lie exactly on this diagonal, and \(T\), the point where the circles touch, also lies on it.
Step 2: Find the length of the diagonal.
The square has side 4 cm, so the diagonal has length $WY = 4\sqrt{2}$ cm (by Pythagoras: $\sqrt{4^2+4^2}$).
Step 3: Express distances along the diagonal in terms of the radii.
Because $O_1$ lies on the diagonal at distance $r_1\sqrt{2}$ from $W$ (the diagonal distance corresponding to a perpendicular offset of $r_1$ from each side), and similarly $O_2$ lies at distance $r_2\sqrt{2}$ from $Y$, the distance between the centers along the diagonal is:
\[ O_1O_2 = WY - r_1\sqrt{2} - r_2\sqrt{2} = 4\sqrt{2} - \sqrt{2}(r_1 + r_2) \]
Step 4: Apply the tangency condition.
Since the circles touch externally, $O_1O_2 = r_1 + r_2$. Substituting $r_1 = 1$:
\[ 4\sqrt{2} - \sqrt{2}(1 + r_2) = 1 + r_2 \]
\[ 4\sqrt{2} - \sqrt{2} - \sqrt{2}\,r_2 = 1 + r_2 \]
\[ 3\sqrt{2} - 1 = r_2(1 + \sqrt{2}) \]
Step 5: Solve and rationalize.
\[ r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{7 - 4\sqrt{2}}{1} = 7 - 4\sqrt{2} \]
This matches the coordinate-geometry result exactly, confirming the diagonal shortcut is valid because both tangent circles nested at opposite corners of a square always have their centers on the same diagonal.
\[ r_2 = \boxed{7 - 4\sqrt{2}\ \text{cm}} \]