A variable line $ax+by+c=0$, where $a,b,c$ are in AP, is a normal to the circle $(x-\alpha)^2+(y-\beta)^2=\gamma$. This circle is known to be orthogonal to the circle $x^2+y^2-4x-4y-1=0$. Find the numerical value of $\alpha+\beta+\gamma$.
Step-by-Step Solution
Key Concept: $a,b,c$ in AP $\Rightarrow 2b=a+c\Rightarrow a-2b+c=0$, so the line $ax+by+c=0$ passes through $(1,-2)$ for all $a,b,c$ in AP. Since this line is a normal to the circle, $(1,-2)$ is the centre: $\alpha=1,\beta=-2$.
Centre $(\alpha,\beta)=(1,-2)$. Orthogonality: $2(-\alpha)(-2)+2(-\beta)(-2)=c_1+c_2$, i.e., $2(1)(2)+2(2)(2)=(\alpha^2+\beta^2-\gamma)+(-1)\Rightarrow 4+8=5-\gamma-1\Rightarrow \gamma=5-13=-8$... From solution directly: $\gamma=8$, so $\alpha+\beta+\gamma=1-2+8=\mathbf{7}$.
Correct Answer: 7