Let centre of a circle C be $\left(\mathrm{\alpha }, \mathrm{\beta }\right)$ and its radius r < 8. Let 3x + 4y = 24 and 3x - 4y = 32 be two tangents and 4x + 3y = 1 be a normal to C. Then $(\mathrm{\alpha }-\mathrm{\beta }+\mathrm{r})$ is equal to

Equation of angular bisectors of the tangents is
$$\begin{gathered} \begin{array}{l}\frac{3\mathrm{x}+4\mathrm{y}-24}{5}=\pm \left(\frac{3\mathrm{x}-4\mathrm{y}-32}{5}\right) \\ \Rightarrow 3\mathrm{x}+4\mathrm{y}-24=3\mathrm{x}-4\mathrm{y}-32 \\ \Rightarrow 8\mathrm{y}=-8\Rightarrow \mathrm{y}=-1 \\ centre \left(\alpha ,\beta \right) lies on \mathrm{y}=-1 \\ \Rightarrow \beta =-1 \\ \sin ce 4x+3y=1 is normal\\ \Rightarrow 4\mathrm{\alpha }+3\mathrm{\beta }=1\Rightarrow 4\mathrm{\alpha }-3=1 \\ \Rightarrow \mathrm{\alpha }=1 \\ (\mathrm{\alpha },\mathrm{\beta })=(1,-1) \\ \mathrm{r}=\frac{|3-4-24|}{5}=5 \\ \therefore \mathrm{\alpha }-\mathrm{\beta }+\mathrm{r}=1+1+5=7\end{array} \end{gathered}$$