Circles
Tangent from External Point
Grade 11
Question:
<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 angle between pair of tangents drawn from the point D to the circle S is:</p>
<p>(a) \(\pi - \tan^{-1}\left(\frac{25}{8}\right)\)</p>
<p>(b) \(\pi - \tan^{-1}\left(\frac{15}{7}\right)\)</p>
<p>(c) \(\pi - \tan^{-1}\left(\frac{15}{8}\right)\)</p>
<p>(d) \(\frac{\pi}{2} - \tan^{-1}\left(\frac{5}{3}\right)\)</p>
Step-by-Step Solution
Key Concept: For tangents from an external point to a circle, the angle between them depends on the distance from point to center and the radius.
<p>The circle touches AB and AD, so the center is at distance r from both these sides. Since AB and AD are perpendicular sides of the rectangle meeting at A, the center lies on the angle bisector from A. With \(|AB| = 32\), \(|AD| = 40\), and \(|BE| = 1\), the radius r can be found. Point D is at distance \(\sqrt{(r-32)^2 + (r-40)^2}\) from center. The angle between tangents from D is \(\pi - \tan^{-1}\left(\frac{15}{8}\right)\).</p>
Correct Answer: c