A circle is tangent to both lines 3x-4y+4=0 and 6x-8y-12=0 and to the x-axis. What is its area and the coordinates of its center?

The distance from 3x-4y+4=0 and 6x-8y-12=0 (which is just 3x-4y-6=0) is |-6-4|/√(3²+(-4)²) which is 2, so the radius is 1 and the area is π.

The centres will be at the intersection of the midline (3x-4y-1=0) and y=1, and the intersection of the midline and y=-1. These centres are (5/3,1) and (-1,-1) respectively.

Graph: https://www.desmos.com/calculator/krbja0pboi

tangent 1: y = ¾x+1

tangent 2: y = ¾x-1.5

The tangent lines are parallel and 2 units apart, so the radius of the circle is 1 and the area of the circle is π units².

Center of circle lies on the line parallel and equidistant to the tangents:

y = ¾x-0.25

Since the circle is tangent to the x-axis, the center lies on y=1 or y=-1.

If the center is on y=1 then

¾x-0.25 = 1

x = 5/3

center (5/3,1)

equation of circle:

(x-5/3)² + (y-1)² = 1

if the center is on y=-1 then

¾x-0.25 = -1

x = -1

center (-1,-1)

equation of circle:

(x+1)² + (y+1)² = 1

Two of the lines are parallel, so there are two congruent circles satisfying the conditions.

The two parallel lines:

3x - 4y + 4 = 0

3x - 4y - 6 = 0 ... This one has been simplified.

distance between the lines

= |4 - (-6)| / √(3² + 4²)

= 2

That is the diameter of the circle. Its radius is 1, and its area is π.

A third parallel line, equidistant from these two must include the center. Simply use the mean of the other two constant terms.

3x - 4y - 1 = 0

These are the two lines parallel to the x-axis and one unit from it:

y = -1

y = 1

The coordinates of the two centers are the solutions to these two systems:

3x - 4y - 1 = 0

y = -1

3x - 4(-1) - 1 = 0

x = -1

(-1, -1)

3x - 4y - 1 = 0

y = 1

3x - 4(1) - 1 = 0

x = 5/3

(5/3, 1)