Circles
Common Chord of Circles
Grade 11
Question:
<p><strong>Paragraph for Questions 634 and 635</strong><br>Let \(PAB\) be a triangle where \(A(1,1)\), \(B(3,3)\) and \(P\) be a variable point such that \(PA^2 + PB^2 = 6\). The locus of point \(P\) is \(S = 0\). From point \(Q(3,7)\), pair of tangents are drawn to the curve \(S = 0\) which touches the curve \(S = 0\) at \(C\) and \(D\). Let \(S_1 = 0\) be the circumcircle of \(\triangle QCD\).</p><p>Equation of common chord of \(S = 0\) and \(S_1 = 0\) is:</p>
<p>(a) \(2x - 3y = 1\)</p>
<p>(b) \(y = x\)</p>
<p>(c) \(3x - 5y = 1\)</p>
<p>(d) \(x + 5y = 13\)</p>
Step-by-Step Solution
Key Concept: The common chord of two circles lies on the line obtained by subtracting their equations (S - S₁ = 0). First, find the locus S using the given condition PA² + PB² = 6, then use the property that the chord of contact from Q to circle S is also the radical axis of S and S₁.
<p><strong>Step 1:</strong> Find locus S using PA² + PB² = 6.</p><p>Let P(x,y). Then (x-1)² + (y-1)² + (x-3)² + (y-3)² = 6</p><p>Expanding: x² - 2x + 1 + y² - 2y + 1 + x² - 6x + 9 + y² - 6y + 9 = 6</p><p>Simplifying: 2x² + 2y² - 8x - 8y + 20 = 6</p><p>∴ S: x² + y² - 4x - 4y + 7 = 0</p><p><strong>Step 2:</strong> Find the chord of contact from Q(3,7) to circle S.</p><p>Chord of contact from (x₁,y₁) to circle x² + y² + 2gx + 2fy + c = 0 is: xx₁ + yy₁ + g(x+x₁) + f(y+y₁) + c = 0</p><p>Here: g = -2, f = -2, c = 7, and (x₁,y₁) = (3,7)</p><p>Chord of contact: 3x + 7y - 2(x+3) - 2(y+7) + 7 = 0</p><p>Simplifying: 3x + 7y - 2x - 6 - 2y - 14 + 7 = 0</p><p>∴ x + 5y - 13 = 0</p><p><strong>Step 3:</strong> Recognize that the common chord of S = 0 and S₁ = 0 is the chord CD (the chord of contact), since S₁ is the circumcircle of △QCD passing through the tangent points C and D.</p><p><strong>Answer:</strong> x + 5y - 13 = 0 (Option D)</p>
Correct Answer: D