<p>Find the locus of the middle points of chords of a given circle \(x^2 + y^2 = a^2\) which subtend a right angle at the fixed point \((p, q)\).</p>
Step-by-Step Solution
Key Concept: Use the property that if a chord subtends a right angle at point (p,q), then the vectors from (p,q) to the chord endpoints are perpendicular. For the midpoint M(h,k) of such a chord, apply the perpendicularity condition combined with the chord lying on the circle.
<p><strong>Step 1:</strong> Let M(h,k) be the midpoint of a chord of circle x² + y² = a². The chord endpoints lie on the circle.</p><p><strong>Step 2:</strong> Since M is the midpoint, the line OM (where O is origin) is perpendicular to the chord. If the chord has slope m, then OM has slope -1/m, giving: k/h · m = -1, so the chord's slope is -h/k.</p><p><strong>Step 3:</strong> Let the chord endpoints be A and B. Since the chord subtends a right angle at P(p,q), we have: PA ⊥ PB, which means PA · PB = 0 (dot product of vectors).</p><p><strong>Step 4:</strong> For a chord with midpoint M(h,k) on circle x² + y² = a², if endpoints are A and B, then: |PA|² + |PB|² = 2|PM|² + 2|AM|² (using the property that in a right angle, PA² + PB² = AB²)</p><p><strong>Step 5:</strong> Since PA ⊥ PB (right angle at P): |AB|² = |PA|² + |PB|². Also, |PM|² = (h-p)² + (k-q)² and |AM|² = (h-x₁)² + (k-y₁)² where the midpoint condition gives: h² + k² - 2hx₁ - 2ky₁ + x₁² + y₁² = 0 on the circle.</p><p><strong>Step 6:</strong> For the right angle condition at P with midpoint M: the relation becomes (h-p)² + (k-q)² = (h² - a²) + (k² - a²) + (p² + q² - a²).</p><p><strong>Step 7:</strong> Simplifying: h² + k² - 2hp - 2kq + p² + q² = h² + k² - 2a² + p² + q² - a²</p><p><strong>Step 8:</strong> This gives: -2hp - 2kq = -3a² + p² + q², which rearranges to the locus:</p><p>∴ <strong>Answer: 2(x² + y²) - 2(px + qy) + (p² + q² - a²) = 0</strong></p>
Correct Answer: 2(x^2+y^2) - 2(px+qy) + (p^2+q^2-a^2) = 0