Circles
Intersection of circles and common chord
Grade 11
Question:
<p>Consider two circles:<br>Equation of \(Q\): \((x-3)^2 + (y+1)^2 = 9\), i.e., \(x^2 + y^2 - 6x + 2y + 1 = 0\)<br>Equation of curve \(P\): \((x-4)^2 + (y-1)^2 = 4\), i.e., \(x^2 + y^2 - 8x - 2y + 13 = 0\)<br>Which of the following are correct?</p>
<p>The two circles intersect at two points.</p>
<p>The equation of common chord is \(2x + 4y - 121 = 0\).</p>
<p>\(x = 6 - 2y\) is the equation of common chord.</p>
<p>The distance between centers is \(\sqrt{5}\).</p>
Step-by-Step Solution
Key Concept: Find the distance between centers and compare with sum/difference of radii to determine the relative position of two circles (external, internal, intersecting, or tangent). Use this to verify geometric properties and relationships between the circles.
<p><strong>Step 1: Extract center and radius from both circles</strong></p><p>Circle Q: (x-3)² + (y+1)² = 9 → Center C₁ = (3, -1), Radius r₁ = 3</p><p>Circle P: (x-4)² + (y-1)² = 4 → Center C₂ = (4, 1), Radius r₂ = 2</p><p><strong>Step 2: Calculate distance between centers</strong></p><p>d = √[(4-3)² + (1-(-1))²] = √[1 + 4] = √5 ≈ 2.236</p><p><strong>Step 3: Determine relative position</strong></p><p>Compare d with r₁ + r₂ and |r₁ - r₂|:</p><p>• r₁ + r₂ = 3 + 2 = 5</p><p>• |r₁ - r₂| = |3 - 2| = 1</p><p>Since 1 < √5 < 5, the circles are <strong>intersecting at two points</strong></p><p><strong>Step 4: Verify statements</strong></p><p>A) Circles intersect at two distinct points ✓ (CORRECT)</p><p>B) Distance between centers = √5 ✓ (CORRECT)</p><p>C) Radical axis exists and is real ✓ (CORRECT for intersecting circles)</p><p>∴ Answer: A, B, C</p>
Correct Answer: A, B, C