The locus of the mid points of the chords of the circle `x^2+y^2+4x-6y-12=0` which subtend an angle of `pi/3`radians at its circumference is:
(A) `(x-2)^2+(y+3)^2=6.25` (B) `(x+2)^2+(y-3)^2=6.25`
(C) `(x+2)^2+(y-3)^2=18.75` (D) `(x+2)^2+(y+3)^2=18.75`

Explanation

We can draw the figure from the given details. Please refer to the video for diagram.

Equation of the given circle is is,

`x^2+y^2+4x-6y-12 = 0`

Comparing this equation with the standard circle equation,

`x^2+y^2+2gx+2fy+c = 0`, we get

`g = 2,f = -3, c = -12`

Now, centre of circle will be `(-g,-f) = (-2,3)`

And, radius AC will be `sqrt(g^2+f^2-c) = sqrt(4+9+12) = 5`

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