<p>The circle S touches the sides AB and AD of the rectangle ABCD and cuts the side DC at a single point F and the side BC at a single point E. If \(|AB| = 32\), \(|AD| = 40\) and \(|BE| = 1\). The radius of circle S is:</p>
Step-by-Step Solution
Key Concept: The circle tangent to two perpendicular sides has its center at equal distance from both; use the constraint that it passes through E to solve for r.
<p>The circle is tangent to sides AB and AD. Place A at origin with AB along x-axis and AD along y-axis. The center is at \((r, r)\) for some radius r. The circle passes through E on BC (where \(|BE| = 1\), so E is at \((32, 39)\)) and through F on DC. Using the distance formula from center to E: \((32-r)^2 + (39-r)^2 = r^2\). Expanding and solving: \(1024 - 64r + r^2 + 1521 - 78r + r^2 = r^2\), giving \(r^2 - 142r + 2545 = 0\). Using the quadratic formula or factoring: \(r = 23\) or \(r = 110.6...\). Since r must be less than 32, we have \(r = 23\).</p>
Correct Answer: b