<p><strong>22.</strong> Through a random point \((p, q)\) on the cartesian plane secants are drawn to the circle \(x^2 + y^2 = r^2\). If the locus of mid-point of the secants to the circle is \(x^2 + 2hxy + y^2 + 2gx + 2fy + c = 0\). Then:</p>
Step-by-Step Solution
Key Concept: The locus of midpoints of all chords of a circle passing through an external point (p,q) forms a circle whose equation is derived using the chord midpoint property: if M(x,y) is a midpoint of a chord, then OM ⊥ to the chord, where O is the center.
<p><strong>Step 1:</strong> For any chord AB of circle x² + y² = r² with midpoint M(x,y), the line OM is perpendicular to AB.</p><p><strong>Step 2:</strong> Since the chord passes through external point P(p,q), we can write: the midpoint M(x,y) satisfies the condition that OM ⊥ PM (property of chord through external point).</p><p><strong>Step 3:</strong> Using the perpendicularity condition: OM · PM = 0 (dot product of direction vectors)</p><p>∴ (x,y) · (x-p, y-q) = 0</p><p>x(x-p) + y(y-q) = 0</p><p>x² + y² - px - qy = 0</p><p><strong>Step 4:</strong> Comparing with x² + 2hxy + y² + 2gx + 2fy + c = 0:</p><p>• Coefficient of xy: 2h = 0 → <strong>h = 0</strong></p><p>• Coefficient of x²: 1 (coefficient of y²): 1</p><p>• 2g = -p → <strong>g = -p/2</strong></p><p>• 2f = -q → <strong>f = -q/2</strong></p><p>• <strong>c = 0</strong></p><p>∴ Answer: B</p>
Correct Answer: B