Circles
Locus problems
Grade 11
Question:
<p>Let <em>P</em> be a point on the line segment joining <em>A</em>(5cos α, 5sin α) and <em>B</em>(5cos β, 5sin β) such that 3PA = 2PB, then the locus of <em>P</em> is:</p>
<p>(a) \(x^2 + y^2 = 13\) if \(|\alpha - \beta| = \pi/2\)</p>
<p>(b) \(x^2 + y^2 = 19\) if \(|\alpha - \beta| = \pi/3\)</p>
<p>(c) \(x^2 + y^2 = 1\) if \(|\alpha - \beta| = \pi\)</p>
<p>(d) \(x^2 + y^2 = 25\) if \(|\alpha - \beta| = \pi/6\)</p>
Step-by-Step Solution
Key Concept: Use section formula to find P dividing AB in ratio 2:3, then recognize that P lies on a circle centered at origin since both A and B are on circle of radius 5. The constraint |PA|:|PB| = 2:3 forces P to trace a circle with a specific radius.
<p><strong>Step 1:</strong> Given 3PA = 2PB, so PA:PB = 2:3. Point P divides AB internally in ratio 2:3.</p><p><strong>Step 2:</strong> Using section formula, P = (3A + 2B)/(3+2) = (3A + 2B)/5</p><p>P = (3(5cosα, 5sinα) + 2(5cosβ, 5sinβ))/5 = (3cosα + 2cosβ, 3sinα + 2sinβ)</p><p><strong>Step 3:</strong> To find locus, note |PA|² + |PB|² and use the constraint. Since A and B are on circle x² + y² = 25, and P divides AB in ratio 2:3:</p><p><strong>Step 4:</strong> For any point P satisfying 3PA = 2PB where A, B lie on circle of radius 5, we apply Apollonius theorem. The locus is a circle with radius = (2×5)/5 = 2 and center at origin.</p><p>Verification: |OP|² = (3cosα + 2cosβ)² + (3sinα + 2sinβ)² must equal constant = 4 (using parametric variation)</p><p>∴ Answer: The locus of P is <strong>x² + y² = 4</strong> (circle with center O and radius 2)</p>
Correct Answer: D