AB=8AB2=64 
⇒(a−b)2+(b+4)2=64 ...(1) Now P divides AB in the ratio 2:1 internally ⇒h=32a+b and k=3−4+b+2 ⇒2a+b=3h … (2) k=3b−2 From equation (2) and (3) ⇒b=3k+2 $\begin{aligned}
& \Rightarrow \quad 2 a=3 h-3 k-2 \
& \Rightarrow \quad a=\frac{3 h-3 k-2}{2}
\end{aligned}Nowbyputtingvalueofaandbinequation\begin{aligned}
& \Rightarrow\left(\frac{3 h-3 k-2}{2}-(3 k+2)\right)^2+(3 k+2+4)^2=64 \
& \Rightarrow\left(\frac{3 h-3 k-2-6 k-4}{2}\right)^2+(3 k+6)^2=64 \
& \Rightarrow(3 h-9 k-6)^2+4(3 k+6)^2=4 \times 64 \
& \Rightarrow 9(h-3 k-2)^2+36(k+2)^2=256 \
& \Rightarrow 9\left(h^2+9 k^2+4-6 h k-4 h+12 k\right) \
& \quad+36\left(k^2+4+4 k\right)=256 \
& \Rightarrow 9\left(h^2+13 k^2+20-6 h k-4 h+28 k\right)=256
\end{aligned}Replacinghbyxandkbyy\begin{aligned}
& \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 y\right)+180-256=0 \
& \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 y\right)-76=0
\end{aligned}Bycomparing\alpha=13, \beta=-6, \gamma=-4\alpha-\beta-\gamma=13+6+4=23$