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 the curve \(q = a\) for some constant \(a \neq 0\), then \(\dfrac{dr}{dp}\) is equal to:</p>
<p>\(\dfrac{p}{4r}\)</p>
<p>\(\dfrac{p}{3r}\)</p>
<p>\(\dfrac{p}{2r}\)</p>
<p>\(\dfrac{p}{r}\)</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. Since q = a (constant), differentiate this constraint with respect to p to find dr/dp.
<p><strong>Step 1: Apply orthogonality condition</strong></p><p>For two circles to be orthogonal at all intersection points, the condition is:</p><p>r² + r² = p² + q²</p><p>⟹ 2r² = p² + q²</p><p><strong>Step 2: Substitute the constraint q = a</strong></p><p>Since the center of C₂ moves along the curve q = a (constant), substitute:</p><p>2r² = p² + a²</p><p><strong>Step 3: Differentiate implicitly with respect to p</strong></p><p>Differentiate both sides with respect to p:</p><p>2·2r·(dr/dp) = 2p + 0</p><p>4r(dr/dp) = 2p</p><p><strong>Step 4: Solve for dr/dp</strong></p><p>dr/dp = 2p/(4r) = p/(2r)</p><p>∴ Answer: C</p>
Correct Answer: C