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.
• B.
• C.
• D.

Answer 1

The correct option is D

Let $P(x_1,y_1)$ be a point in the locus
Radius of $S\equiv x^2+y^2+4x-6y+9si{n}^2\alpha +13co{s}^2\alpha =0$
$r=\sqrt{(-2{)}^2+(-3{)}^2-9si{n}^2\alpha -13co{s}^2\alpha }=\sqrt{13(1-co{s}^2\alpha )-9si{n}^2\alpha }=2sin\alpha$
Length of the tangent from $P(x_1,y_1)$ to S = 0 is $\sqrt{S_{11}}=\sqrt{x_1^2+y_1^2+4x_1-6y_1+9si{n}^2\alpha +13co{s}^2\alpha }$
Since $2\alpha$ is the angle between the tangent drawn from P to S = 0
we have $tan\alpha =\frac{r}{\sqrt{s_{11}}}\Rightarrow S_{11}.ta{n}^2\alpha =r^2\Rightarrow (x_1^2+y_1^2+4x_1-6y_1+9si{n}^2\alpha +13co{s}^2\alpha )ta{n}^2\alpha =4si{n}^2\alpha \Rightarrow (x_1^2+y_1^2+4x_1-6y_1+9+4co{s}^2\alpha )\frac{si{n}^2\alpha }{co{s}^2\alpha }=4si{n}^2\alpha \Rightarrow (x_1^2+y_1^2+4x_1-6y_1+9)se{c}^2\alpha =0\Rightarrow x_1^2+y_1^2+4x_1-6y_1+9=0Locusof(x_1,y_1)is{x}^2+y^2+4x-6y+9=0$