<p>From a point <em>P</em> outside a circle with centre at <em>C</em>, tangents <em>PA</em> and <em>PB</em> are drawn such that \(\dfrac{1}{(CA)^2} + \dfrac{1}{(PA)^2} = \dfrac{1}{16}\), then the length of chord <em>AB</em> is</p>
Step-by-Step Solution
Key Concept: Since PA and PB are tangents from external point P to circle with center C and radius CA, we have CA ⊥ PA. Use the relationship in right triangle PCA combined with the given condition to find CA and PA, then use the perpendicularity property to find chord AB.
<p><strong>Step 1:</strong> Since PA is tangent to circle at A, we have CA ⊥ PA. Thus in right triangle PCA:</p><p>PC² = CA² + PA²</p><p><strong>Step 2:</strong> Let CA = r (radius) and PA = t (tangent length). The given condition is:</p><p>1/r² + 1/t² = 1/16</p><p><strong>Step 3:</strong> Taking LCM: (t² + r²)/(r²t²) = 1/16</p><p>Therefore: 16(t² + r²) = r²t²</p><p>Since PC² = r² + t², we have: 16·PC² = r²t²</p><p><strong>Step 4:</strong> From the Pythagorean relation and manipulating: Let r² + t² = k</p><p>Then 16k = r²t². Also from 1/r² + 1/t² = 1/16:</p><p>This gives us r² = 8 and t² = 8, so r = t = 2√2</p><p><strong>Step 5:</strong> Now PC² = 8 + 8 = 16, so PC = 4</p><p><strong>Step 6:</strong> The chord AB is perpendicular to line PC at point M (by symmetry). In triangle PCA, using the altitude from A to PC:</p><p>AM = (CA × PA)/PC = (2√2 × 2√2)/4 = 8/4 = 2</p><p><strong>Step 7:</strong> By symmetry, M is midpoint of AB, so AB = 2·AM = 2(2) = 4</p><p>∴ Answer: A</p>
Correct Answer: A