Circles
Area of triangle related to circle
Grade 11
Question:
<p>The circle \(x^2 + y^2 = 1\) cuts the x-axis at P and Q. Another circle with centre at Q and variable radius intersects to first circle at R above the x-axis and the line segment PQ at S. The maximum area of the triangle QSR is</p>
<p>(a) \(\dfrac{2}{9}\)</p>
<p>(b) \(\dfrac{5\sqrt{2}}{7}\)</p>
<p>(c) \(\dfrac{4\sqrt{3}}{9}\)</p>
<p>(d) \(\dfrac{\sqrt{2}}{13}\)</p>
Step-by-Step Solution
Key Concept: The triangle QSR has maximum area when SR is perpendicular to QS (making it a right triangle at S), and we must use the constraint that R lies on the unit circle while S lies on the diameter PQ.
<p><strong>Step 1:</strong> Set up coordinates. The unit circle x² + y² = 1 intersects the x-axis at P(-1,0) and Q(1,0). Point S lies on segment PQ, so S = (s, 0) where -1 ≤ s ≤ 1.</p><p><strong>Step 2:</strong> Point R is on the unit circle above the x-axis and also on the circle centered at Q with radius QS. Let R = (cos θ, sin θ) where θ ∈ (0,π). Then QS = |s - 1| and QR = radius = |s - 1|.</p><p><strong>Step 3:</strong> Calculate QR: QR² = (cos θ - 1)² + sin² θ = cos² θ - 2cos θ + 1 + sin² θ = 2 - 2cos θ = 2(1 - cos θ). So QR = √[2(1 - cos θ)].</p><p><strong>Step 4:</strong> From constraint QR = QS: √[2(1 - cos θ)] = 1 - s, giving s = 1 - √[2(1 - cos θ)].</p><p><strong>Step 5:</strong> Area of triangle QSR = ½ · QS · h, where h is the height from R to the x-axis = sin θ. So Area = ½ · √[2(1 - cos θ)] · sin θ.</p><p><strong>Step 6:</strong> Let u = 1 - cos θ, so sin θ = √[2u - u²]. Area = ½ · √(2u) · √[2u - u²] = ½√[2u(2u - u²)] = ½√[2u²(2 - u)] = ½u√[2(2-u)].</p><p><strong>Step 7:</strong> Maximize f(u) = u√(2 - u) where u ∈ (0,2). Taking derivative: f'(u) = √(2-u) + u · (-1)/(2√(2-u)) = (2(2-u) - u)/(2√(2-u)) = (4 - 3u)/(2√(2-u)). Setting f'(u) = 0: u = 4/3.</p><p><strong>Step 8:</strong> At u = 4/3: f(4/3) = (4/3)√(2/3) = (4/3) · √2/√3 = 4√2/(3√3) = 4√6/9. Maximum area = ½ · √2 · 4√6/9 = 2√12/9 = 4√3/9.</p><p>∴ Answer: C</p>
Correct Answer: C