Circles
Intersection of circles
Grade 11

Question:

<p>If the circles \(x^2 + y^2 - 16x - 20y + 164 = r^2\) and \((x - 4)^2 + (y - 7)^2 = 36\) intersect at two distinct points, then:</p>
<p>\(r &gt; 11\)</p>
<p>\(0 &lt; r &lt; 1\)</p>
<p>\(r = 11\)</p>
<p>\(1 &lt; r &lt; 11\)</p>

Step-by-Step Solution

Key Concept: Convert the first circle to standard form to find its center and radius, then use the condition that two circles intersect at two distinct points when the distance between centers satisfies: |r₁ - r₂| < d < r₁ + r₂.
<p><strong>Step 1:</strong> Convert the first circle to standard form by completing the square.</p><p>x² + y² - 16x - 20y + 164 = r²</p><p>(x² - 16x + 64) + (y² - 20y + 100) + 164 - 64 - 100 = r²</p><p>(x - 8)² + (y - 10)² = r²</p><p>Center C₁ = (8, 10), radius = r</p><p><strong>Step 2:</strong> Identify the second circle's center and radius.</p><p>(x - 4)² + (y - 7)² = 36</p><p>Center C₂ = (4, 7), radius = 6</p><p><strong>Step 3:</strong> Find the distance between centers.</p><p>d = √[(8-4)² + (10-7)²] = √[16 + 9] = √25 = 5</p><p><strong>Step 4:</strong> Apply the condition for two distinct intersection points.</p><p>For two distinct intersection points: |r - 6| < 5 < r + 6</p><p>From |r - 6| < 5: -5 < r - 6 < 5 → 1 < r < 11</p><p>From 5 < r + 6: r > -1</p><p>Combined with r > 0 (radius must be positive): 1 < r < 11</p><p>∴ Answer: D</p>
Correct Answer: D

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