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  • The angles 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 \cos ^2 \alpha+13 \sin ^2 \alpha=0 \text { is } 2 \alpha \text {. }}" src="https://

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

Option: 1

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


Option: 2

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


Option: 3

\mathrm{x^2+y^2+4 x-6 y-\left(4-13 \cot ^2 \alpha\right)=0}


Option: 4

\mathrm{x^2+y^2+4 x-6 y+\left(13-4 \cot ^2 \alpha\right)=0}


Answers (1)

best_answer

The coordinates of the centre and radius of the given circle are (–2, 3) and \sqrt{4+9-9 \cos ^2 \alpha-13 \sin ^2 \alpha}=2 \cos \alpha respectively.
Let the coordinates of P be (h, k).

\mathrm{\text { Clearly, } C P \text { bisects } \angle T P T^{\prime} \text {. }}′.
\mathrm{\therefore \quad \angle C P T=\angle C P T^{\prime}=\alpha}
Now, in DCPT, we have

\mathrm{\begin{aligned} & \sin \alpha=\frac{C T}{C P} \Rightarrow \sin \alpha=\frac{2 \cos \alpha}{\sqrt{(h+2)^2+(k-3)^2}} \\ & \Rightarrow(h+2)^2+(k-3)^2=4 \cot ^2 \alpha \\ & \Rightarrow h^2+k^2+4 h-6 k+\left(13-4 \cot ^2 \alpha\right)=0 \end{aligned}}

\mathrm{\text { Hence, the locus of }(h, k) \text { is } x^2+y^2+4 x-6 y+\left(13-4 \cot ^2 \alpha\right)=0}

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rishi.raj

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