If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0

Given both lines are parallel \& tangent to the circle.

Then distance between these two lines must be the diameter of the circle. Lines:

$3 x-4 y+4=0$

$3 x-4 y-3.5=0$ (Equating the co-efficient of the equation)

Distance (between 2 || lines) $=\left(\frac{|c-p|}{\sqrt{a^{2}+b^{2}}}\right)$ for the two given equations,

$a x+b y+c=0 \& a x+b y+p=0$

Distance $=\left(\frac{|4-(-3.5)|}{\sqrt{3^{2}+(-4)^{2}}}\right)=\left(\frac{|7.5|}{\sqrt{9+16}}\right)=\frac{7.5}{5}=1.5$ units

Radius $=\frac{\text { Diameter }}{2}=\frac{1.5}{2}$

$=0.75$ units.

Hence, the radius of the circle is $0.75=3 / 4$ units.