Question

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

The center of the circle
$x^2+y^2+4x-6y+9si{n}^2\alpha +13co{s}^2\alpha =0$
is C(-2, 3) and its radius is
$\sqrt{2^2+(-3{)}^2-9si{n}^2\alpha -13co{s}^2\alpha }\Rightarrow \sqrt{4+9-9si{n}^2\alpha -13co{s}^2\alpha }=|2sin\alpha |$


Let P(h, k) be any point on the locus. Then $\angle APC=\alpha$
From the diagram.
$sin\alpha =\frac{AC}{PC}=\frac{2sin\alpha }{\sqrt{(h+2{)}^2+(k-3{)}^2}}or(h+2{)}^2+(k-3{)}^2=4or{h}^2+k^2+4h-6k+9=0$
Thus, the required equation of the locus is
$x^2+y^2+4x-6y+9=0$