The angle between a pair of tangents drawn from a point P to the circle x 2- y 2-4 x -6 y -9 sin 2-13 cos 2-0 is 2 - The equation of the locus of the point P isA- x2-y2-4 x-6 y-9-0- x2-y2-4 x-6 y-4-0C- x2-y2-4 x-6 y-4-0D- x2-y2-4 x-6 y-9-0

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Angle between pair of tangents

The angle bet...

Question

The angle between a pair of tangents drawn from a point P to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 s i n^{2} α + 13 c o s^{2} α = 0 i s 2 α .$ The equation of the locus of the point P is

$x^{2} + y^{2} + 4 x - 6 y + 4 = 0$

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$x^{2} + y^{2} + 4 x - 6 y - 9 = 0$

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$x^{2} + y^{2} + 4 x - 6 y - 4 = 0$

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$x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

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Solution

The correct option is D
$x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

Let PA and PB be the tangents drawn from the point P(h, k) to the given circle with centre C(-2, 3). So, that
$∠ A P B = 2 α a n d ∠ A P C = ∠ C P B = α ∠ P A C = ∠ P B C = 90^{\circ} F r o m t r i a n g l e P C A, \Rightarrow s i n α = \frac{C A}{C P} a n d C A = \sqrt{4 + 9 - (9 s i n^{2} α + 13 c o s^{2} α)} = 2 s i n α \Rightarrow C P = 2. \Rightarrow 4 = h^{2} + k^{2} + 4 h - 6 k + 13.$
The locus of P(h, k) is
$x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

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The angle between a pair of tangents drawn from a point P to the circle x2+y2+4x−6y+9sin2α+13cos2α=0 is 2α. The equation of the locus of the point P is

The angle between a pair of tangents drawn from a point P to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 s i n^{2} α + 13 c o s^{2} α = 0 i s 2 α .$ The equation of the locus of the point P is