The lines 3x-4y+4=0 and 6x-8y-7=0 are tangents to the same circle the radius of the circle is

1.11/5
2.3/2
3.3/4
4.11/10

3x-4y+4=0.........(1)
6x-8y-7=0
2(3x-4y-7/2)=0
3x-4y-7/2=0..........(2)
(1) and (2) are parallel
distance between
(1) and (2) =l4+7/2l/sqrt(9+16)
=15/2*5
=3/2
radius=3/2/2
=3/4

3x-4y+4=0 6x-8y-7=0
y=0.75x+1 y=0.75x-0.875

These two lines have the same slope so they are parallel to each other. By looking at their different y=intercepts, you can see that they are 1.875 apart in the y direction which is the hypotenuse of the 3-4-5 triangle form by this vertical distance, a perpendicular line to one of the lines, and a connecting portion of one of the lines. You need to solve for the length of the perpendicular line connecting the two lines, which is the second longest side of this triangle. Using the ratio of these two sides:

4/5 = d/(15/8)
d=3/2
r=3/4

The formula to find the distance from a point (x0,y0) to a line Ax+By+C=0 is
d = | A(x0) + B(y0)+C| / sqrt(A^2 + B^2)

step 1) Make sure these are 2 paralle lines.
3x - 4Y + 4 = 0 <-- (1)
3x - 4y - 7/2 = 0 <-- (2)

step 2) pick a point on one of the lines.
From line (1), we pick the point (0,1). This point satisfies line (1) equation.

step 3) Use the formula to find the distance from (0,1) to line (2),
d = | 3(0) - 4(1) - 7/2| / sqrt(3^2 + 4^2)
d = 3/2