SOLUTION: Find the equation of the circle tangent to the line 4x – 3y + 12=0 at (-3, 0) and also tangent to the line 3x + 4y –16 = 0 at (4, 1).

We could do this the long way using formulas for the distance
of a point to a line, but we don't need to.

The given lines happen to be perpendicular, and the two radii
are perpendicular to the lines, and the radii are equal, so
the figure has to be a square. The x-intercept of the line that
goes up to the right is (-3,0), which is a corner of the square.
The upper vertex of the square is 3 units right and 4 units up
from the left vertex of the square.  Therefore the bottom vertex
of the square has to be 3 units left and 4 units down from the
right vertex of the square, so the bottom vertex of the square
is (1,-3) and that is the center of the circle. The two radii
are hypotenuses of 3-4-5 right triangles so the radius is 5.
So the equation of the circle is

(x-1)²+(y+3)²=25

Edwin