Circles
Locus of points
Grade 11
Question:
<p>If <strong>A</strong>(<i>a</i>, 0) and <strong>B</strong>(−<i>a</i>, 0) are two fixed points and a point P moves such that ∠APB = 90°, then locus of P is</p>
<p>(a) \(x^2 + y^2 = 2a^2\)</p>
<p>(b) \(x^2 + y^2 = a^2\)</p>
<p>(c) \(x^2 + y^2 + 2a^2 = 0\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the property that if an angle inscribed in a circle is 90°, then the chord subtending it is a diameter. Apply Pythagoras theorem to find the locus.
<p><strong>Step 1:</strong> Let coordinates of point P be $(h, k)$.</p><p><strong>Step 2:</strong> Given, ∠APB = 90°</p><p><strong>Step 3:</strong> By Pythagoras theorem, we have $AB^2 = PA^2 + PB^2$</p><p><strong>Step 4:</strong> $(2a)^2 = (h-a)^2 + k^2 + (h+a)^2 + k^2$</p><p><strong>Step 5:</strong> $4a^2 = h^2 + a^2 - 2ah + h^2 + a^2 + 2ah + 2k^2$</p><p><strong>Step 6:</strong> $4a^2 = 2h^2 + 2a^2 + 2k^2$</p><p><strong>Step 7:</strong> $2a^2 = 2h^2 + 2k^2$</p><p><strong>Step 8:</strong> $h^2 + k^2 = a^2$</p><p><strong>Step 9:</strong> Therefore, $x^2 + y^2 = a^2$ is the required locus.</p><p>∴ Answer is (b).</p>
Correct Answer: b