Circles
Orthogonal circles and locus
Grade 11
Question:
<p>If a circle passes through the point (<em>a</em>, <em>b</em>) and cuts the circle \(x^2 + y^2 = p^2\) orthogonally, then the equation of the locus of its centre is</p>
<p>\(x^2 + y^2 - 3ax - 4by + (a^2 + b^2 - p^2) = 0\)</p>
<p>\(2ax + 2by - (a^2 - b^2 + p^2) = 0\)</p>
<p>\(x^2 + y^2 - 2ax - 3by + (a^2 - b^2 - p^2) = 0\)</p>
<p>\(2ax + 2by - (a^2 + b^2 + p^2) = 0\)</p>
Step-by-Step Solution
Key Concept: Two circles cut orthogonally when the tangents at their intersection points are perpendicular, which occurs when 2g₁g₂ + 2f₁f₂ = c₁ + c₂. Use this condition combined with the constraint that the variable circle passes through (a,b) to find the locus of its center.
<p><strong>Step 1:</strong> Let the variable circle have center (h, k) and radius r. Its equation is: (x - h)² + (y - k)² = r²</p><p><strong>Step 2:</strong> The given circle x² + y² = p² has center (0, 0) and radius p. Expanding the variable circle: x² + y² - 2hx - 2ky + (h² + k² - r²) = 0</p><p><strong>Step 3:</strong> For two circles S₁: x² + y² + 2g₁x + 2f₁y + c₁ = 0 and S₂: x² + y² + 2g₂x + 2f₂y + c₂ = 0 to cut orthogonally: 2g₁g₂ + 2f₁f₂ = c₁ + c₂</p><p><strong>Step 4:</strong> Applying orthogonality condition: Our circle has g = -h, f = -k, c = h² + k² - r². The given circle has g = 0, f = 0, c = -p². Therefore: 2(-h)(0) + 2(-k)(0) = (h² + k² - r²) + (-p²)</p><p>This gives: 0 = h² + k² - r² - p², so r² = h² + k² - p²</p><p><strong>Step 5:</strong> Since the variable circle passes through (a, b): (a - h)² + (b - k)² = r²</p><p><strong>Step 6:</strong> Substituting r² = h² + k² - p²: (a - h)² + (b - k)² = h² + k² - p²</p><p><strong>Step 7:</strong> Expanding: a² - 2ah + h² + b² - 2bk + k² = h² + k² - p²</p><p>Simplifying: a² + b² - 2ah - 2bk = -p²</p><p><strong>Step 8:</strong> Replacing (h, k) with (x, y) for the locus: 2ax + 2by = a² + b² + p²</p><p>∴ Answer: <strong>2ax + 2by = a² + b² + p²</strong> (or equivalently ax + by = (a² + b² + p²)/2)</p>
Correct Answer: D