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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+12 x-16 y+36 \sin ^2 \alpha+100 \cos ^2 \alpha=0\, \, is \, \, 2 \alpha.}" src="https://entran

The angle between a pair of tangents drawn from a point P to the circle \mathrm{x^2+y^2+12 x-16 y+36 \sin ^2 \alpha+100 \cos ^2 \alpha=0\, \, is \, \, 2 \alpha.} The equation of the locus of the point P is

Option: 1

\mathrm{x^2+y^2-12 x-16 y+64=0}


Option: 2

\mathrm{x^2+y^2-12 x-16 y-36=0}


Option: 3

\mathrm{x^2+y^2-12 x-16 y-64=0}


Option: 4

\mathrm{x^2+y^2+12 x-16 y+36=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(-6,8). So that

\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 $\triangle PCA, we have

\mathrm{ \begin{aligned} \sin \alpha & =\frac{C A}{C P} \text { and } C A=\sqrt{36+64-\left(36 \sin ^2 \alpha+100 \cos ^2 \alpha\right)} \\ & =8 \sin \alpha \\ \Rightarrow & C P=8 . \\ \Rightarrow & 64=h^2+k^2+12 h-16 k+100 \end{aligned} }

Thus, the locus of P(h, k) is

\mathrm{ x^2+y^2+12 x-16 y+36=0 \text {. } }

Posted by

SANGALDEEP SINGH

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