Circles
Greatest distance from a point to a circle
Grade None
Question:
<p>Let \(S \equiv x^2 + y^2 - 4x - 2y - 20\). If \(P(10, 7)\) lies outside the circle \(S = 0\), the greatest distance of point \(P\) from \(S = 0\) is:</p>
<p>3 unit</p>
<p>4 unit</p>
<p>5 unit</p>
<p>\(\sqrt{64+36} - 5\) unit</p>
Step-by-Step Solution
Key Concept: The greatest distance from an external point P to a circle equals the distance from P to the center plus the radius. This occurs along the line joining P and the center, on the far side of the circle.
<p><strong>Step 1:</strong> Rewrite circle equation in standard form.</p><p>S ≡ x² + y² - 4x - 2y - 20 = 0</p><p>(x² - 4x + 4) + (y² - 2y + 1) - 4 - 1 - 20 = 0</p><p>(x - 2)² + (y - 1)² = 25</p><p>Center C = (2, 1), Radius r = 5</p><p><strong>Step 2:</strong> Verify P(10, 7) is outside the circle.</p><p>Distance from P to C: PC = √[(10-2)² + (7-1)²] = √[64 + 36] = √100 = 10</p><p>Since PC = 10 > r = 5, point P is outside the circle ✓</p><p><strong>Step 3:</strong> Find greatest distance from P to circle S = 0.</p><p>Greatest distance = PC + r = 10 + 5 = 15</p><p>∴ Answer: D</p>
Correct Answer: D