Question

As shown in the figure, circle C1 with center O1 and radius r1 touches the square VWXY at points P and Q while circle C2 with center O2 and radius r2 touches the square VWXY at points R and S. The two circles touch each other at T.
Given r1 = 1 cm and VY = VW = 4 cm, r2 = cm.

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

Hint:

Place the square on coordinate axes, write both circle centers in terms of r1 and r2 using the tangency conditions, then equate the distance between centers to r1+r2.

Answer 1

The Correct Option is C

Step 1: Set up coordinates.
Place the square VWXY with W = (0,0), X = (4,0), V = (0,4), Y = (4,4), since the square has side length 4 cm. Circle C1 sits in the bottom-left corner touching sides VW and WX, circle C2 sits in the top-right corner touching sides YX and VY.
Step 2: Locate the centers.
Circle C1 is tangent to VW (x=0) and WX (y=0), so O1 = (r1, r1) = (1, 1). Circle C2 is tangent to YX (x=4) and VY (y=4), so O2 = (4-r2, 4-r2).
Step 3: Apply the external tangency condition.
Since the two circles touch each other externally at T, the distance between their centers equals r1 + r2.\[ O_1O_2 = \sqrt{2}(3 - r_2) = 1 + r_2 \]
Step 4: Solve for r2. \[ 3\sqrt{2} - \sqrt{2}r_2 = 1 + r_2 \implies r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}} = 7 - 4\sqrt{2} \]
Step 5: Conclude. \[ \boxed{r_2 = 7 - 4\sqrt{2}\ \text{cm}} \]