The angle between a pair of tangents from a point P to the circe `x^2 + y^2+ 4 x-6y + 9 sin^2 alpha + 13 cos^2 alpha =0` is `2alpha`. Find the equation of the locus of the point P.

Options

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

D

Explanation

The coordinates of the centre and radius of the given circle are (-2, 3) and `sqrt(4+9-9sin^(2)alpha-13cos^(2)alpha)=2sin alpha` respectively .
Let the coordinates of CP be (h, k).
Clearly, CP bisects `angleTPT'=2alpha`
`:. angleCPT=angleCPT'=alpha`
Now, in `DeltaCPT`, we have

`sin alpha=(CT)/(CP)`
`rArr sin alpha=(2sin alpha)/(sqrt((h+2)^(2)+(k-3)^(2)))`
`rArr (h+2)^(2)+(k-3)^(2)=4rArrh^(2)+k^(2)+4h-6k+9=0`
Hence, the locus of (h, k) is `x^(2)+y^(2)+4x-6y+9=0`