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
Both circle centers lie on the diagonal joining the two corners the circles hug. Use the tangency condition that the distance between centers equals the sum of the radii.
Step 1: Set up coordinates for the square. Let corner \(W\) be the origin, with the square's sides along the axes. Since the square has side 4 cm, \(W = (0,0)\), \(X = (4,0)\), \(V = (0,4)\), and \(Y = (4,4)\). Side \(VW\) lies along the y-axis and side \(WX\) lies along the x-axis. Step 2: Locate the center \(O_1\). Circle \(C_1\) is tangent to sides \(VW\) and \(WX\), the two sides meeting at corner \(W\). A circle tangent to both axes has its center at a distance equal to its radius from each axis, so \(O_1 = (r_1, r_1) = (1,1)\). Step 3: Locate the center \(O_2\). Circle \(C_2\) is tangent to sides \(VY\) (the top side, \(y = 4\)) and \(XY\) (the right side, \(x = 4\)), the two sides meeting at corner \(Y\). By the same reasoning, its center is at a perpendicular distance \(r_2\) from each of those sides, so \(O_2 = (4 - r_2,\ 4 - r_2)\). Step 4: Use the external tangency condition. Since the two circles touch each other at \(T\), the distance between their centers equals the sum of their radii: \(O_1O_2 = r_1 + r_2\). Compute \(O_1O_2\): \[ O_1O_2 = \sqrt{(4-r_2-1)^2 + (4-r_2-1)^2} = \sqrt{2}\,(3 - r_2) \] Step 5: Solve for \(r_2\). Setting the two expressions equal: \[ \sqrt{2}\,(3 - r_2) = 1 + r_2 \] \[ 3\sqrt{2} - \sqrt{2}\,r_2 = 1 + r_2 \] \[ 3\sqrt{2} - 1 = r_2\,(1 + \sqrt{2}) \] \[ r_2 = \frac{3\sqrt{2} - 1}{1 + \sqrt{2}} \] Step 6: Rationalize the denominator. Multiply 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 7: Sanity-check against the other options. Numerically \(r_2 = 7 - 4\sqrt{2} \approx 1.34\) cm, a small positive radius consistent with the figure. Option (A), \(4 - 3\sqrt{2} \approx -0.24\), is negative, which is impossible for a radius. Options (B) and (D), \(1 + 2\sqrt{2} \approx 3.83\) and \(5 + 3\sqrt{2} \approx 9.24\), would push \(O_2\) past the center line of the square (a valid \(r_2\) must be less than 2 for both circles to fit without crossing the diagonal midpoint), so they are also inconsistent with the tangency configuration shown. \[ r_2 = \boxed{7 - 4\sqrt{2}\ \text{cm}} \]
