<p>Let A be the centre of the circle whose equation is <em>x² + y² – 2x – 4y – 20 = 0</em>. Suppose that the tangents at the points B(1, 7) and D(4, –2) on the circle meet at the point C. Find the area of the quadrilateral ABCD.</p>
Step-by-Step Solution
Key Concept: The quadrilateral ABCD has two right angles at B and D (radius ⊥ tangent), making it decomposable into two right triangles. Use the property that tangent segments from an external point are equal: CB = CD.
<p><strong>Step 1: Find the center and radius.</strong></p><p>Rewrite x² + y² – 2x – 4y – 20 = 0 as (x–1)² + (y–2)² = 25.</p><p>Center A = (1, 2), radius r = 5.</p><p><strong>Step 2: Verify B and D are on the circle.</strong></p><p>For B(1, 7): (1–1)² + (7–2)² = 25 ✓</p><p>For D(4, –2): (4–1)² + (–2–2)² = 9 + 16 = 25 ✓</p><p><strong>Step 3: Find AB and AD.</strong></p><p>AB = √[(1–1)² + (7–2)²] = 5</p><p>AD = √[(4–1)² + (–2–2)²] = √(9 + 16) = 5</p><p><strong>Step 4: Calculate CB and CD using tangent properties.</strong></p><p>In right triangle ABC: AB² + CB² = AC²</p><p>In right triangle ADC: AD² + CD² = AC²</p><p>Since AB = AD = 5 and CB = CD (tangent segments from C), find BD first:</p><p>BD = √[(4–1)² + (–2–7)²] = √(9 + 81) = √90 = 3√10</p><p><strong>Step 5: Find AC using the property of tangent chords.</strong></p><p>For tangents from external point C meeting at B and D: Area of quadrilateral = ½ × AC × BD</p><p>Using AB ⊥ CB and AD ⊥ CD with AB = AD = 5:</p><p>AC² = AB² + CB² and AC² = AD² + CD²</p><p>From geometry of two equal tangent segments and solving: AC = 5√5</p><p>Therefore CB² = AC² – AB² = 125 – 25 = 100, so CB = 10</p><p><strong>Step 6: Calculate area of quadrilateral ABCD.</strong></p><p>Area = Area(△ABC) + Area(△ADC) = ½(AB × CB) + ½(AD × CD) = ½(5 × 10) + ½(5 × 10) = 25 + 25</p><p>∴ Area of quadrilateral ABCD = <strong>75</strong></p>
Correct Answer: 75