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

Answer 1

The correct option is D $x^2+y^2+4x-6y+9=0$

$x^2+y^2+4x-6y+9{\sin }^2\alpha +13{\cos }^2\alpha =0\Rightarrow (x+2{)}^2+(y-3{)}^2=9-9{\sin }^2\alpha +4-13{\cos }^2\alpha \Rightarrow (x+2{)}^2+(y-3{)}^2=9{\cos }^2\alpha -13{\cos }^2\alpha +4\Rightarrow (x+2{)}^2+(y-3{)}^2=4-4{\cos }^2\alpha \Rightarrow (x+2{)}^2+(y-3{)}^2=4{\sin }^2\alpha \therefore \text{Centre of the circle}C(-2,3),\text{and radius}r=2\sin \alpha$

According to the image
$\sin \alpha =\frac{AC}{PC}P{C}^2{\sin }^2\alpha =4{\sin }^2\alpha$
$(h+2{)}^2+(k-3{)}^2=4$
$\therefore \text{locus of P is}{x}^2+y^2+4x-6y+9=0$