If the equation of the circle throught the points of intersection of the circles `x^(2)+y^(2)-4x-6y-12=0`
and `x^(2)+y^(2)+"Ax"+"By"+C=0` and intersecting the circle `x^(2)+y^(2)-2x-4=0`
orthogonally is `x^(2)+y^(2)+"Ax"+"By"+C=0`, then find teh value of `(A+B+C)`.

Explanation

Equation of the family of circles through the points of intersection of the circles
`x^(2)+y^(2)-4x-6y-12=0" and "x^(2)+y^(2)+6x+4y-12=0` is given by
`x^(2)+y^(2)-4x-6y-12+lambda(x^(2)+y^(2)+6x+4y-12)=0`
i.e.`" "x^(2)+y^(2)+((6lamda-4))/(1+lamda)x+((4lamda-6))/(1+lamda)y-12=0" "...(1)`
Since one of the circle given by (1) intersects the circle `x^(2)+y^(2)-2x=4`
orthogonally, we have, (for that value of `lambda`),
`2.((3lambda-2)/(1+lambda))(-1)+0=-12-4implies3lambda-2=8(1+lambda)implies5lambda=-10,lambda=-2`
Equation of the required circle is : `x^(2)+y^(2)+16x+14y-12=0`
`implies" "A=16,B=14" and "C=-12`
Hence `A+B+C=16+14+6(-12)=18`