Here the given lines are parallel.
i.e., $3 x+4 y+4=0 \& 3 x+4 y-\frac{7}{2}=0$
Diameter $=$ distance between parallel tangents
$$
=\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{\left|4+\frac{7}{2}\right|}{\sqrt{(3)^2+(4)^2}}=\frac{15}{10}=\frac{3}{2}
$$
$\Rightarrow r=\frac{1}{2}$ diameter $=\frac{1}{2} \cdot \frac{3}{2}=\frac{3}{4}$
Let $(-\mathrm{a},-\mathrm{a})$ be the centre of the circle $r=$ perpendicular distance from $(-a,-a)$ on $3 x-4 y+8=0$
Let $(-a,-a)$ bet he centre of the circle
$$
\Rightarrow r=\frac{-3 a+4 a+8}{\sqrt{(3)^2+(-4)^2}}=\frac{a+8}{5} \text {, }
$$
also $r=a$