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

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

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

let (h,k) be any point P and
`angleAPC=alpha,anglePAC-pi//2`
That is , triangle APC is a right angled triangle.
`therefore sinalpha=(AC)/(PC)=(2sinalpha)/(sqrt(h+2^(2)+(k-3)^(2)))`
`rArr(h+2)^(2)+(k-3)^(2)=4`
`rArrh^(2)+4+4h+k^(2)+9-6k=4`
`rArrh^(2)+k^(2)+4h-6k+9=0`
Thus, required equation of the locus is
`x^(2)+y^(2)+4x-6y+9=0`