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Let C be the circle x^2+y^2+4x-6y-3=0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x=6 is

CAT · 2023 · Quant Slot 1
Question:
Let C be the circle x^2+y^2+4x-6y-3=0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x=6 is

Options

(6,4)
(6,8)
(6,3)
(6,6)
Answer: (6,3)

Explanation:
Circle: x^2 + y^2 + 4x − 6y − 3 = 0 Complete squares: (x + 2)^2 + (y − 3)^2 = 16 Center = (−2, 3) Radius r = 4 For a point P outside a circle, angle between tangents = 2θ, where sinθ = r/d Given angle = 60° ⇒ 2θ = 60° ⇒ θ = 30° So sin30° = 1/2 = r/d = 4/d Therefore d = 8 So locus of such points (from where tangents make 60°) is a circle centered at (−2, 3) with radius 8: (x + 2)^2 + (y − 3)^2 = 64 We need point where this locus touches the line x = 6: Distance between x = 6 and center x = −2 = |6 − (−2)| = 8 This equals the radius, so the line x = 6 is tangent to the locus. The tangent point has same y-coordinate as center (3). Thus point of tangency = (6, 3) Correct option: (6, 3)

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