Circles
Orthogonal circles and locus
Grade 11
Question:
<p>If a circle passes through the point \((a, b)\) and cuts the circle \(x^2 + y^2 = 4\) orthogonally, then the locus of its centre is</p>
<p>\(2ax + 2by + (a^2 + b^2 + 4) = 0\)</p>
<p>\(2ax + 2by - (a^2 + b^2 + 4) = 0\)</p>
<p>\(2ax - 2by + (a^2 + b^2 + 4) = 0\)</p>
<p>\(2ax - 2by - (a^2 + b^2 + 4) = 0\)</p>
Step-by-Step Solution
Key Concept: When two circles cut orthogonally, the condition is 2g₁g₂ + 2f₁f₂ = c₁ + c₂. For a circle passing through (a,b) and cutting x² + y² = 4 orthogonally, use this orthogonality condition combined with the constraint that (a,b) lies on the variable circle.
<p><strong>Step 1:</strong> Let the circle with center (h, k) and radius r be: (x - h)² + (y - k)² = r²</p><p><strong>Step 2:</strong> The given circle x² + y² = 4 has center (0, 0) and radius 2.</p><p><strong>Step 3:</strong> For orthogonal circles, the condition is: 2(0)(h) + 2(0)(k) = 4 + r², which gives r² = -4 (this form needs correction). The correct orthogonality condition is: h² + k² = r² + 4</p><p><strong>Step 4:</strong> Since (a, b) lies on the circle: (a - h)² + (b - k)² = r²</p><p><strong>Step 5:</strong> Expanding: a² + b² - 2ah - 2bk + h² + k² = r²</p><p><strong>Step 6:</strong> Substituting r² = h² + k² - 4 from orthogonality condition: a² + b² - 2ah - 2bk + h² + k² = h² + k² - 4</p><p><strong>Step 7:</strong> Simplifying: a² + b² - 2ah - 2bk = -4</p><p><strong>Step 8:</strong> Rearranging with (h, k) as variables: 2ah + 2bk = a² + b² + 4</p><p>∴ The locus is: <strong>2ax + 2by = a² + b² + 4</strong> (or equivalently ax + by = (a² + b² + 4)/2)</p>
Correct Answer: B