Circles
Maximum distance
Grade 11
Question:
<p>If a point \(P\) has coordinates \((0, -2)\) and \(Q\) is any point on the circle, \(x^2 + y^2 - 5x - y + 5 = 0\), then the maximum value of \((PQ)^2\) is</p>
<p>\(\dfrac{25 + \sqrt{6}}{2}\)</p>
<p>\(8 + 5\sqrt{3}\)</p>
<p>\(14 + 5\sqrt{3}\)</p>
<p>\(\dfrac{47 + 10\sqrt{6}}{2}\)</p>
Step-by-Step Solution
Key Concept: The maximum distance from an external point to any point on a circle equals the distance from the point to the circle's center plus the circle's radius. Square this maximum distance to get (PQ)².
<p><strong>Step 1:</strong> Rewrite the circle equation in standard form.</p><p>x² + y² - 5x - y + 5 = 0</p><p>(x² - 5x + 25/4) + (y² - y + 1/4) + 5 - 25/4 - 1/4 = 0</p><p>(x - 5/2)² + (y - 1/2)² = 25/4 + 1/4 - 5 = 26/4 - 20/4 = 6/4 = 3/2</p><p>Center C = (5/2, 1/2), Radius r = √(3/2) = √6/2</p><p><strong>Step 2:</strong> Find distance PC from P(0, -2) to center C(5/2, 1/2).</p><p>PC² = (5/2 - 0)² + (1/2 - (-2))² = 25/4 + (5/2)² = 25/4 + 25/4 = 50/4 = 25/2</p><p>PC = √(25/2) = 5/√2 = 5√2/2</p><p><strong>Step 3:</strong> Maximum value of PQ occurs when Q lies on the line joining P and C, on the far side of C from P.</p><p>PQ(max) = PC + r = 5√2/2 + √6/2 = (5√2 + √6)/2</p><p><strong>Step 4:</strong> Calculate (PQ)²(max).</p><p>(PQ)²(max) = [(5√2 + √6)/2]² = (5√2 + √6)²/4 = (50 + 6 + 2·5√2·√6)/4 = (56 + 10√12)/4 = (56 + 20√3)/4 = 14 + 5√3</p><p>∴ Answer: D (14 + 5√3 or equivalent form)</p>
Correct Answer: D