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Tangents are drawn to the circle x² + y² = 1 at its intersection points (distinct) with the circle x² + y² + (lambda - 3)x + (2*lambda + 2)y + 2 = 0. As lambda varies, the locus of the intersection point of the pair of tangents is a straight line. Find the slope of that straight line.

1. -3
2. -1/3
3. 1/2
4. 2

Correct answer: 2

Solution

The common chord of the two circles is the radical axis: subtract to get (lambda-3)x + (2*lambda+2)y + 3 = 0, or -(3-lambda)x - (2*lambda+2)y = 3. The pole of this chord w.r.t. x²+y²=1 (i.e., the intersection of tangents at the two common points) satisfies: hx+ky = 1 is the same line as (lambda-3)x + (2*lambda+2)y + 3 = 0 (rewritten as [(lambda-3)/(-3)]x + [(2*lambda+2)/(-3)]y = 1). So h = (lambda-3)/(-3) = (3-lambda)/3, k = (2*lambda+2)/(-3). Eliminate lambda: from h: lambda = 3 - 3h. Substitute into k: k = -(2(3-3h)+2)/3 = -(6-6h+2)/3 = -(8-6h)/3 = (6h-8)/3. So 3k = 6h - 8 → 6h - 3k = 8 → 2h - k = 8/3. Locus: 2x - y = 8/3, slope = 2.

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