Circles
Locus of a point
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 3<em>PA</em> = 2<em>PB</em> 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: P divides AB in ratio 2:3, so use section formula to find coordinates of P, then eliminate parameters α and β to get the locus equation. The key is recognizing that both A and B lie on a circle of radius 5 centered at origin.
<p><strong>Step 1:</strong> From 3PA = 2PB, we get PA:PB = 2:3, so P divides AB internally in ratio 2:3.</p><p><strong>Step 2:</strong> Using section formula, if P = (x,y), then:</p><p>x = (3·5cos α + 2·5cos β)/(2+3) = (3cos α + 2cos β)</p><p>y = (3·5sin α + 2·5sin β)/(2+3) = (3sin α + 2sin β)</p><p><strong>Step 3:</strong> Square and add:</p><p>x² + y² = 9cos²α + 12cos α cos β + 4cos²β + 9sin²α + 12sin α sin β + 4sin²β</p><p>x² + y² = 9(cos²α + sin²α) + 4(cos²β + sin²β) + 12(cos α cos β + sin α sin β)</p><p>x² + y² = 9 + 4 + 12cos(α - β)</p><p><strong>Step 4:</strong> Since P lies on segment AB where A and B are on circle x² + y² = 25, and α - β varies, the minimum value occurs when cos(α - β) = -1 and maximum when cos(α - β) = 1:</p><p>Minimum: x² + y² = 13 - 12 = 1</p><p>Maximum: x² + y² = 13 + 12 = 25</p><p>∴ The locus is the annular region: <strong>1 ≤ x² + y² ≤ 25</strong> or the circle <strong>x² + y² = 25</strong> depending on answer options provided.</p>
Correct Answer: D