AB is any chord of the circle `x^2+y^2-6x-8y-11=0` which subtends an angle `pi/2` at `(1,2)`. If locus of midpoint of AB is a circle `x^2+y^2-2ax-2by-c=0`; then find the value of `(a+b+c)`.

Explanation

`r=6`

`h=(3+6costheta+3+6costheta)/2`

`k=(4+6sintheta+4+6sintheta)/2`

`(2h-6)/6=costheta+cosphi`

`costheta+cosphi=(h-3)/3-(1)`

`(2k-8)/6=sintheta+sinphi`

`sintheta+sinphi=(k-4)/3-(2)`

`1+1+2(cos(theta-phi))=((h-3)/3)^2+((k-4)/3)^2`

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