Circles
Tangents from external point
Grade 11
Question:
<p><b>Paragraph for Question nos. 616 and 617</b><br>Consider a circle \(S: x^2 + y^2 - 6x - 4y - 3 = 0\) with centre \(C\) and \(P\) be the point \((-1, -1)\). Also \(PA\) and \(PB\) are tangents drawn to the circle \(S\).</p><p>The area of the quadrilateral \(PACB\) is equal to:</p>
<p>(a) 12</p>
<p>(b) 24</p>
<p>(c) \(3\sqrt{15}\)</p>
<p>(d) \(4\sqrt{15}\)</p>
Step-by-Step Solution
Key Concept: For a quadrilateral formed by two tangents from external point P to circle with center C, the area equals the product of the tangent length and the radius, since PACB comprises two right triangles sharing the radius as height. Use the formula: Area = tangent length × radius.
<p><strong>Step 1:</strong> Rewrite circle equation in standard form:</p><p>x² + y² - 6x - 4y - 3 = 0</p><p>(x - 3)² + (y - 2)² = 9 + 4 + 3 = 16</p><p>Centre C = (3, 2), Radius r = 4</p><p><strong>Step 2:</strong> Find distance PC:</p><p>PC = √[(−1−3)² + (−1−2)²] = √[16 + 9] = √25 = 5</p><p><strong>Step 3:</strong> Find tangent length PA using right triangle PAC:</p><p>Since PA is tangent, ∠PAC = 90°</p><p>PA² = PC² − r² = 25 − 16 = 9</p><p>PA = 3</p><p><strong>Step 4:</strong> Calculate area of quadrilateral PACB:</p><p>PACB consists of two right triangles: △PAC and △PBC (congruent)</p><p>Area(PACB) = 2 × Area(△PAC) = 2 × (½ × PA × r)</p><p>= 2 × (½ × 3 × 4) = 2 × 6 = 12</p><p>∴ Answer: D (Area = 12)</p>
Correct Answer: D