Radical axis of circle $x^2+y^2+2 \alpha x+2 \beta y+c=0$ and $x^2+y^2+\frac{3}{2} x+4 y+c=0$ is
$\begin{aligned}
& \left(2 \alpha-\frac{3}{2}\right) x+(2 \beta-4) y=0 \\
& (4 \alpha-3) x+4(\beta-2) y=0 \\
& (4 \alpha-3) x+(\beta-8) y=0 \text { touches the circle } \\
& x^2+y^2+2 x+2 y+1=0 \\
& \therefore \quad 1=\left|\frac{(4 \alpha-3)(-1)+(3-8)(-1)}{\sqrt{(4 \alpha-3)^2+(4 \beta-8)^2}}\right| \\
& \Rightarrow \quad(4 \alpha-3)^2+(4 \mathrm{~b}-8)^2=(4 \alpha+4 \beta-11)^2 \\
& \Rightarrow \quad 16 \alpha^2-24 \alpha+9+16 \beta^2-64 \beta+64 \\
& =16 \alpha^2+16 \beta^2+121+32 \alpha \beta-88 \alpha-88 \beta \\
& \Rightarrow 32 \alpha \beta-88 \alpha+24 \alpha-88 \beta+64 \beta=9+64-121 \\
& 32 \alpha \beta-64 \alpha-24 \beta=-48 \Rightarrow 4 \alpha \beta-8 \alpha-3 \beta=-6 \\
& \Rightarrow \quad 4 \alpha \beta-8 \alpha-3 \beta+10=-6+10=4 \\
&
\end{aligned}$