Mathematics | VITEEE 2017 | Discussion Page

The angle between a pair of tangants drawn from a point T to the circle

$x^{2}+y^{2}+4x-6y+9\sin^{2} \alpha+13\cos^{2} \alpha =0$ is 2α

The equation of the locus of the point T is

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

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

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

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

Option D

Radius of circle

$=\sqrt{4+9-9\sin^{2}\alpha-13\cos^{b}\alpha}=2 \begin{vmatrix}\sin \alpha\end{vmatrix}$

If T be (h,k) then

$\tan\alpha=\frac{2 \begin{vmatrix}\sin \alpha\end{vmatrix}}{\sqrt{h^{2}+k^{2}+4h-6k+9\sin^{2}\alpha+ 13\cos^{2}\alpha}}$

$\Rightarrow{h^{2}+k^{2}+4h-6k+9\sin^{2}\alpha+ 13\cos^{2}\alpha}=4\cos^{2}\alpha$

$\Rightarrow{h^{2}+k^{2}+4h-6k+9=0}$

∴ Locus of T is ${x^{2}+y^{2}+4x-6y+9=0}$

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