Question

The angle between a pair of tangents drawn from a point P to the circle $x^2+y^2+4x-6y+9si{n}^2\alpha +13co{s}^2\alpha =0is2\alpha .$ 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 D

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

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
$\angle APB=2\alpha and\angle APC=\angle CPB=\alpha \angle PAC=\angle PBC={90}^{\circ }FromtrianglePCA,\Rightarrow sin\alpha =\frac{CA}{CP}andCA=\sqrt{4+9-(9si{n}^2\alpha +13co{s}^2\alpha )}=2sin\alpha \Rightarrow CP=2.\Rightarrow 4=h^2+k^2+4h-6k+13.$
The locus of P(h, k) is
$x^2+y^2+4x-6y+9=0$