Circles
Orthogonal circles and locus
Grade 11
Question:
<p><strong>Paragraph for Questions 576 and 577:</strong><br>Let \(C_1 : x^2 + y^2 = r^2\) and \(C_2 : (x-p)^2 + (y-q)^2 = r^2\) be 2 circles with radius \(r\) \((r > 0)\) and have \(n\) points of intersection, \((x_i, y_i)\) for \(i \in \{1, 2, \ldots, n\}\). If \(C_1\) and \(C_2\) are orthogonal at all points of intersection, then:<br><br>As we move the centre of \(C_2\) along \(p + bq = 0\) for some constant \(b \neq 0\), then \(\dfrac{dr}{dq}\) is equal to:</p>
<p>\(\dfrac{b^2 q + q}{r}\)</p>
<p>\(\dfrac{b^2 q + q}{2r}\)</p>
<p>\(\dfrac{b^2 q + q}{3r}\)</p>
<p>\(\dfrac{b^2 q + q}{4r}\)</p>
Step-by-Step Solution
Key Concept: For orthogonal circles, the condition is that the sum of squares of radii equals the square of distance between centers: r² + r² = p² + q². As the center moves along the constraint p + bq = 0, differentiate this orthogonality condition with respect to q to find dr/dq.
<p><strong>Step 1:</strong> Write the orthogonality condition for two circles with equal radii r: The circles are orthogonal when r² + r² = p² + q², which gives 2r² = p² + q².</p><p><strong>Step 2:</strong> Apply the constraint on center of C₂: The center (p, q) moves along p + bq = 0, so p = -bq.</p><p><strong>Step 3:</strong> Substitute the constraint into orthogonality condition: 2r² = (-bq)² + q² = b²q² + q² = q²(b² + 1).</p><p><strong>Step 4:</strong> Differentiate both sides with respect to q: 2·2r·(dr/dq) = 2q(b² + 1), which gives 4r(dr/dq) = 2q(b² + 1).</p><p><strong>Step 5:</strong> Solve for dr/dq: dr/dq = q(b² + 1)/(2r).</p><p><strong>Step 6:</strong> Express in terms of known quantities. From Step 3: q²(b² + 1) = 2r², so q(b² + 1) = 2r²/q. Therefore dr/dq = (2r²/q)/(2r) = r/q.</p><p>∴ Answer: B</p>
Correct Answer: B