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

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

4

Explanation

The center of the circle
`x^(2)+y^(2)+4x-6y+9 sin ^(2) alpha+13 cos^(2) alpha =0`
is `C(-2,3)` and its radius is
`sqrt(4+9-9 sin ^(2) alpha-13 cos ^(2) alpha)=|2 sin alpha|`

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