1.

The locus of the mid points of the chords of the circle `x^2+y^2+4x-6y-12=0` which subtends an angle of `pi/3` radians at its centre is

Answer» Equation of the circle,
`x^2+y^2+4x-6y -12 = 0`
Here, `g = 2, f = -3, c = -12`
`:.` Center of the circle will be `(-2,3)`.
Radius of the circle ` = sqrt(g^2+f^2-c) = sqrt(4+9+12) = 5`
Let `AB` is the chord and `C(h,k)` is the midpoint of `AB`.
We have to find locus of point `C`.
Let `O(-2,3)` is the center of the circle.
Then, `/_AOB = pi/3 =>/_AOC = 1/2/_AOB = pi/6`.
Now, `(OC)/(OA) = cos (pi/6)`
`=>OC = 5**sqrt3/2 = 5sqrt3/2`...(As OA = radius)
`:. sqrt((h+2)^2+(k-3)^2) = (5sqrt3)/2`
Squaring both sides,
`=>((h+2)^2+(k-3)^2) = ((5sqrt3)/2)^2`
`=>(h+2)^2+(k-3)^2 = 75/4`
Replacing `(h,k)` with `(x,y)`,
`(x+2)^2+(y-3)^2 = 75/4`, which is the required locus.


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