Question

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)`.

Answer 1

`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`
`2+2cos90=(h^2+k^2-6h-8k+25)/9`
`18=h^2+k^2-6h-8k+25`
`x^2+y^2-6x-8y+17=0`
`2a=6`
`a=3`
`2b=8`
`b=4`
`c=-17`
a+b+c=3+4-17=-10.