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 circles sit in opposite right-angle corners of the square, so their centers lie a distance equal to their own radius from each of the two sides meeting at that corner; set the center-to-center distance equal to \(r_1 + r_2\) for external tangency.

Answer 1

The Correct Option is C

Step 1: Set up coordinates using the corners the circles are tangent to.
Place \(W\) (the bottom-left corner) at the origin, with side \(WX\) along the positive x-axis and side \(WV\) along the positive y-axis. Since \(\overline{VW} = \overline{VY} = 4\) cm, the square has side length 4 cm, so the corners are \(W(0,0)\), \(X(4,0)\), \(V(0,4)\), \(Y(4,4)\).

Step 2: Locate the center \(O_1\).
Circle \(C_1\) touches the square at \(P\) (on side \(VW\)) and \(Q\) (on side \(WX\)), the two sides that meet at corner \(W\). A circle tangent to both sides of a right angle has its center at a distance equal to its radius from each side. So \(O_1 = (r_1, r_1) = (1, 1)\).

Step 3: Locate the center \(O_2\).
Circle \(C_2\) touches the square at \(R\) (on side \(XY\)) and \(S\) (on side \(VY\)), the two sides that meet at corner \(Y\). By the same reasoning, its center is a distance \(r_2\) from each of these two sides, measured inward from \(Y(4,4)\). So \(O_2 = (4 - r_2,\ 4 - r_2)\).

Step 4: Apply the external tangency condition.
The circles touch each other at \(T\), so they are externally tangent, meaning the distance between their centers equals the sum of their radii:
\[ O_1O_2 = r_1 + r_2 \]
Compute \(O_1O_2\) using the distance formula:
\[ O_1O_2 = \sqrt{(4 - r_2 - 1)^2 + (4 - r_2 - 1)^2} = \sqrt{2(3 - r_2)^2} = (3 - r_2)\sqrt{2} \]
(taking \(3 - r_2 > 0\), which must hold since \(r_2\) is a small radius inside a 4 cm square).

Step 5: Solve for \(r_2\).
\[ (3 - r_2)\sqrt{2} = r_1 + r_2 = 1 + r_2 \]
\[ 3\sqrt{2} - r_2\sqrt{2} = 1 + r_2 \]
\[ 3\sqrt{2} - 1 = r_2 + r_2\sqrt{2} = r_2(1 + \sqrt{2}) \]
\[ r_2 = \frac{3\sqrt{2} - 1}{1 + \sqrt{2}} \]
Rationalize by multiplying top and bottom by \((\sqrt{2} - 1)\):
\[ r_2 = \frac{(3\sqrt{2} - 1)(\sqrt{2} - 1)}{(1 + \sqrt{2})(\sqrt{2} - 1)} = \frac{6 - 3\sqrt{2} - \sqrt{2} + 1}{2 - 1} = 7 - 4\sqrt{2} \]

Why the other options are wrong:
\(4 - 3\sqrt{2}\) is negative (since \(3\sqrt{2} \approx 4.24\)), which cannot be a radius.
\(1 + 2\sqrt{2} \approx 3.83\) cm is far too large for a circle tucked into that corner while staying tangent to \(C_1\) at that position.
\(5 + 3\sqrt{2}\) is even larger and clearly impossible inside a 4 cm square.

Final Answer:
\[ \boxed{r_2 = 7 - 4\sqrt{2} \text{ cm} \approx 1.34 \text{ cm}} \]