Step 1: Observe the given lines.
The lines are
\[
3x-4y+4=0
\]
and
\[
6x-8y-7=0
\]
Divide the second equation by \(2\):
\[
3x-4y-\frac72=0
\]
Thus, the two lines are parallel.
Step 2: Use the property of tangents to a circle.
If two parallel lines are tangents to a circle, then the distance between the lines equals the diameter of the circle.
Hence,
\[
\text{Radius}=\frac12\times(\text{distance between the lines})
\]
Step 3: Find the distance between the parallel lines.
The lines are:
\[
3x-4y+4=0
\]
and
\[
3x-4y-\frac72=0
\]
Distance between parallel lines
\[
ax+by+c_1=0
\]
and
\[
ax+by+c_2=0
\]
is
\[
\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}
\]
Thus,
\[
d=
\frac{\left|4-\left(-\frac72\right)\right|}{\sqrt{3^2+(-4)^2}}
\]
\[
=
\frac{\left|4+\frac72\right|}{\sqrt{9+16}}
\]
\[
=
\frac{\frac{15}{2}}{5}
\]
\[
=\frac32
\]
Step 4: Find the radius.
Since diameter
\[
=\frac32,
\]
radius is
\[
r=\frac12\times\frac32
\]
\[
=\frac34
\]
Step 5: Final conclusion.
Therefore,
\[
\boxed{\frac34}
\]