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  • The angle between a pair of tangents drawn from a point P to the circle <img alt="\mathrm{x^2+y^2+4 x-6 y+9 \sin ^2 \alpha+13 \cos ^2 \alpha=0 \, \, is \, \, 2 \alpha.}" src="https://entrancec

The angle between a pair of tangents drawn from a point P to the circle \mathrm{x^2+y^2+4 x-6 y+9 \sin ^2 \alpha+13 \cos ^2 \alpha=0 \, \, is \, \, 2 \alpha.} The equation of the locus of the point P is

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

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

\mathrm{ \begin{aligned} & \angle A P B=2 \alpha \text { and } \angle A P C=\angle C P B=\alpha \\ & \angle P A C=\angle P B C=90^{\circ} \end{aligned} }

In  \mathrm{\triangle P C A}, we have

\mathrm{ \begin{aligned} & \sin \alpha=\frac{C A}{C P} \text { and } C A=\sqrt{4+9-\left(9 \sin ^2 \alpha+13 \cos ^2 \alpha\right)} \\\\ & =2 \sin \alpha \\\\ & \Rightarrow C P=2 . \\\\ & \Rightarrow 4=h^2+k^2+4 h-6 k+13 \end{aligned} }

Thus, the locus of P(h, k) is \mathrm{x^2+y^2+4 x-6 y+9=0.}

Posted by

Devendra Khairwa

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