A variable line $ax+by+c=0$ where a, b, c are in AP, is normal to a circle $(x-\alpha)^2+(y-\beta)^2=\gamma$, which is orthogonal to the circle $x^2+y^2-4x-4y-1=0$. The value of $\alpha+\beta+\gamma$ is _______.
Attempt
Normal crosses the centre of circle so_$a\alpha+b\beta+c=0$.
Also AP condition led to: $2a\alpha+a\beta+c\beta+2c=0$
Orhtogonal cicles condition gave equation : $15{\alpha}^2+15{\beta}^2- 8\alpha- 8\beta+32\alpha\beta+\gamma=0$
But from here I am not able to do anything.
Any suggestions?
The orthogonality condition on "http://mathworld.wolfram.com/OrthogonalCircles.html" (this is just simplificaiton of the result $r_1^2+r_2^2=d^2$ from the same site, where $r_1^2=g^2+f^2-c,r_2^2=g'^2+f'^2-c'\ and\ d^2=(g-g')^2+(f-f')^2$) would give $4\alpha +4\beta +1=\alpha^2+\beta^2-\gamma-1\Rightarrow (\alpha-2)^2+(\beta-2)^2=\gamma+9$
The straight line(which is variable) can be written by using 2b-a=c into the form a(x-1)+b(y+2)=0, i.e. it always passes through 1 and -2, but it must in every case be normal to the circle, i.e., must pass through its center and it would only be possible if $\alpha=1$ and $\beta=-2$, putting it into the orthogonality condition would yield $\gamma=8$. Thus $\alpha+\beta+\gamma=7$