<p>For any real k, the circle \(x^2 + y^2 + 2kx + 2ky - 8 = 0\) passes through two fixed points A and B. Locus of the point of intersection of the tangents to the circle at A and B is:</p>
Step-by-Step Solution
Key Concept: The locus of the intersection of tangents at two fixed points lies on the chord of contact, which is the radical axis of the family of circles.
<p>Rewrite the circle equation as \(x^2 + y^2 - 8 + 2k(x + y) = 0\). For all k, the fixed points A and B satisfy \(x^2 + y^2 - 8 = 0\) and \(x + y = 0\) simultaneously. Solving: from \(x + y = 0\), we get \(y = -x\), so \(x^2 + x^2 = 8\), giving \(x = \pm 2\). The fixed points are \((2, -2)\) and \((-2, 2)\). The locus of the pole (intersection of tangents at A and B) with respect to any circle in the family is the radical axis, which is \(x + y = 0\).</p>
Correct Answer: C