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:

Place each circle's center at a distance equal to its own radius from the two square sides it touches, then use the external tangency condition (center distance = r1+r2) with Pythagoras.

Answer 1

The Correct Option is C

Step 1: Set up coordinates for the square.
Let the square VWXY have side length 4 cm, since \(\overline{VY} = \overline{VW} = 4\) cm. Place corner W at the origin \((0,0)\), corner X at \((4,0)\), corner V at \((0,4)\), and corner Y at \((4,4)\), so the sides WX and WV run along the axes.

Step 2: Locate the center of circle \(C_1\).
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 a distance equal to its radius from each side. So the center \(O_1 = (r_1, r_1) = (1, 1)\), since \(r_1 = 1\) cm.

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

Step 4: Use the tangency condition between the circles.
The two circles touch each other at T, meaning they are externally tangent. For externally tangent circles, the distance between centers equals the sum of the radii:
\[ O_1O_2 = r_1 + r_2 \]
Compute \(O_1O_2\) using the distance formula:
\[ O_1O_2 = \sqrt{[(4-r_2) - 1]^2 + [(4-r_2)-1]^2} = \sqrt{2}\,(3 - r_2) \]

Step 5: Solve the equation.
\[ \sqrt{2}\,(3 - r_2) = 1 + r_2 \]
\[ 3\sqrt{2} - \sqrt{2}\,r_2 = 1 + r_2 \]
\[ 3\sqrt{2} - 1 = r_2(1+\sqrt{2}) \]
\[ r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}} \]
Rationalize by multiplying numerator and denominator by \((\sqrt{2}-1)\):
\[ r_2 = \frac{(3\sqrt{2}-1)(\sqrt{2}-1)}{(1+\sqrt{2})(\sqrt{2}-1)} = \frac{6 - 3\sqrt{2} - \sqrt{2} + 1}{2-1} = 7 - 4\sqrt{2} \]

Step 6: Check why other options are wrong.
Option (A), \(4-3\sqrt{2}\), works out to about \(-0.24\), which is negative and cannot be a radius. Option (B), \(1+2\sqrt{2}\), is about 3.83, too large to fit as the smaller circle drawn beside \(C_1\) inside the square. Option (D), \(5+3\sqrt{2}\), is far larger than the square's own side. Only option (C) gives a sensible, positive radius smaller than the square.

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