Question

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

• A. $x^2+y^2+4x-6y+4=0$
• B. $x^2+y^2+4x-6y-9=0$
• C. $x^2+y^2+4x-6y-4=0$
• D. $x^2+y^2+4x-6y+9=0$

Answer 1

The correct option is A $x^2+y^2+4x-6y+4=0$

$LetthecentrebeO,pointsoncirclefrom$ $where\tan gentsaredrawnisA,Bandpointofinter\sec tionof\tan gentisP{x}^2+y^2+4x-6y+9{\sin }^{2}x+13{\cos }^{2}x=0Centreofthecircle=\left(-2,3\right)Radiusofcircle=\sqrt{{\left(-2\right)}^2+{\left(3\right)}^2-9{\sin }^{2}x-13{\cos }^{2}x}=\sqrt{13-9{\sin }^{2}x-13CI\ast {\sin }^{2}x}=\sqrt{4{\sin }^{2}x}=2\sin x=OA2xistheanglebetween\tan gents\sin x=\frac{OA}{OP}O\left(-2,3\right)P\left(h,k\right)\sin x=\frac{2\sin x}{\sqrt{{\left(h+2\right)}^2+{\left(k-3\right)}^2}}\Rightarrow {\left(h+2\right)}^2+{\left(k-3\right)}^2=4\Rightarrow h^2+k^2+4h-6k+9x=4\Rightarrow h^2+k^2+4h-6k+9=0focusofpointPis{x}^2+y^2+4x-6y+9=0OptionAiscorrectanswer.$