Circles
Position of Point and Distance from Circle
Grade 11

Question:

<p>The least and the greatest distances of the point (10, 7) from the circle \(x^2 + y^2 - 4x - 2y - 20 = 0\) are</p>
<p>(a) 10, 5</p>
<p>(b) 15, 20</p>
<p>(c) 12, 16</p>
<p>(d) 5, 15</p>

Step-by-Step Solution

Key Concept: Convert circle to standard form to find centre and radius, then use distance formula to find minimum (PC - r) and maximum (PC + r) distances.
<p><strong>Step 1:</strong> Rewrite the circle equation in standard form: \(x^2 + y^2 - 4x - 2y - 20 = 0\)</p><p><strong>Step 2:</strong> Complete the square: \((x-2)^2 + (y-1)^2 = 4 + 1 + 20 = 25\)</p><p><strong>Step 3:</strong> Centre C = (2, 1), radius r = 5</p><p><strong>Step 4:</strong> Distance from P(10, 7) to C(2, 1): \(PC = \sqrt{(10-2)^2 + (7-1)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\)</p><p><strong>Step 5:</strong> Minimum distance = PC - r = 10 - 5 = 5</p><p><strong>Step 6:</strong> Maximum distance = PC + r = 10 + 5 = 15</p><p>∴ Answer is (d) 5, 15</p>
Correct Answer: D

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