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

Since circle C1 is tangent to the two sides meeting at corner W, its center O1 lies on the diagonal from W to Y. Circle C2's center O2 also lies on that same diagonal.
The full diagonal WY has length $4\sqrt{2}$ cm. Center O1 is at distance $r_1\sqrt{2}=\sqrt{2}$ cm from W along the diagonal. Center O2 is at distance $r_2\sqrt{2}$ cm from Y.
\[ O_1O_2 = 4\sqrt{2} - \sqrt{2} - r_2\sqrt{2} = \sqrt{2}(3-r_2) \]
Setting this equal to $r_1+r_2=1+r_2$ and solving gives $r_2 = 7-4\sqrt{2}$ cm.