The equation of the circle (Radius >0) whose every tangent is perpendicular to exactly one of the member of the family of the $x+y-2+\lambda (7x-3y-4)=0$ at the point of contact and the circle touches only one member of the family $(2x-3y)+\mu (x-y-1)=0$, is given by $x^2+y^2+ax+by+c=0$, then $|a+b+c|$=________.
My approach is as follow. The point of intersection of $x+y-2+\lambda=0$ and $7x-3y-4=0$ is (1,1)
The point of intersection of $2x-3y=0$ and $x-y-1=0$ is (3,2).
We assume that (1,1) is the point where the tangent intersect at $90^o$.
and (3,2) refer to the point where the line touches the circle viz. tangent
Let $(h,k)$ be the centre of the circle anf $r$ be the radius $(h-3)^2+(k-2)^2=r^2$
The locus of the point of intersection of perpendicular tangents refer to the director circle.
If $x^2+y^2=r^2$ refer to the circle then its director circle is $x^2+y^2=2r^2$
Hence $r^2=(3-1)^2+(2-1)^2=5$.
Locus of the centre of the circle is $(x-3)^2+(y-2)^2=5^2$
How do we find the circle.
HINT.
The circle is centred at $(1,1)$ and passes through $(3,2)$.