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.
Show Hint
Place the square on coordinates so \(O_1\) sits at distance \(r_1\) from two adjacent sides and \(O_2\) sits at distance \(r_2\) from the other two adjacent sides. Use the fact that the circles touch, so the distance between \(O_1\) and \(O_2\) equals \(r_1+r_2\), and solve the resulting equation for \(r_2\).
Step 1: Understanding the Concept:
Circle \(C_1\) touches two adjacent sides of the square, the left side and the bottom side, so its center \(O_1\) must be at a distance \(r_1\) from each of those two sides. In the same way, circle \(C_2\) touches the other two adjacent sides, the top side and the right side, so its center \(O_2\) must be at a distance \(r_2\) from each of those sides. Since the two circles touch each other, the distance between their centers equals the sum of their radii.
Step 2: Key Formula or Approach:
Place the square on coordinates with \(W = (0,0)\), \(X = (4,0)\), \(V = (0,4)\), \(Y = (4,4)\), since the side length is 4 cm. Then \(O_1 = (r_1, r_1)\) and \(O_2 = (4 - r_2, 4 - r_2)\). The circles touching each other externally at \(T\) means:
\[ O_1O_2 = r_1 + r_2 \]
Step 3: Detailed Explanation:
With \(r_1 = 1\), \(O_1 = (1,1)\). The difference between the two centers along each axis is the same: \(3 - r_2\). So the squared distance between the centers is:
\[ O_1O_2^2 = (3-r_2)^2 + (3-r_2)^2 = 2(3-r_2)^2 \]
Setting this equal to \((r_1+r_2)^2 = (1+r_2)^2\):
\[ 2(3-r_2)^2 = (1+r_2)^2 \]
Taking the positive square root of both sides, since \(r_2 < 3\) makes both \(3-r_2\) and \(1+r_2\) positive:
\[ \sqrt{2}\,(3-r_2) = 1+r_2 \]
Step 4: Solve for \(r_2\).
Expand and collect the \(r_2\) terms on one side:
\[ 3\sqrt{2} - \sqrt{2}\,r_2 = 1 + r_2 \]
\[ 3\sqrt{2} - 1 = r_2(1+\sqrt{2}) \]
\[ r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}} \]
Multiply the top and bottom by \((\sqrt{2}-1)\) to remove the surd from the bottom:
\[ r_2 = \frac{(3\sqrt{2}-1)(\sqrt{2}-1)}{(1+\sqrt{2})(\sqrt{2}-1)} = \frac{6 - 3\sqrt{2} - \sqrt{2} + 1}{2-1} = 7 - 4\sqrt{2} \]
Step 5: Sanity check the number.
\(7 - 4\sqrt{2} \approx 7 - 5.657 = 1.343\) cm. This is positive and less than 3 cm, so the value is consistent with the figure, where circle \(C_2\) is clearly larger than circle \(C_1\) but still fits inside the square.
Final Answer:
The radius of circle \(C_2\) works out to \(7 - 4\sqrt{2}\) cm.
\[ \boxed{r_2 = 7 - 4\sqrt{2} \text{ cm}} \]
