The circle is tangent to the line 4x +3y = 4 at the point P(4,-4) and the center is on the line x-y = 7.

You need to know the formula for the circle? It seems difficult at first, but it won't be so bad. First, we need to describe a line that is perpendicular to 4x + 3y = 4 and passes through point P.

If we have a line:

Ax + By = C

Then the perpendicular line will be

Bx - Ay = D

3x - 4y = D
3 * 4 - 4 * (-4) = D
12 + 16 = D
28 = D

3x - 4y = 28

Now, let's find when 3x - 4y = 28 intersects x - y = 7

x - y = 7
x = 7 + y

3 * x - 4 * y = 28
3 * (7 + y) - 4y = 28
21 + 3y - 4y = 28
-y = 7
y = -7

x = 7 + y
x = 7 - 7
x = 0

So the center of the circle is at (0 , -7)

Now, find the distance from (0 , -7) to (4 , -4)

r^2 = (-4 - (-7))^2 + (4 - 0)^2
r^2 = (-4 + 7)^2 + 4^2
r^2 = 3^2 + 4^2
r^2 = 9 + 16
r^2 = 25
r = 5


So our circle is centered at (0 , -7) and it has a radius of 5. It's formula would be:

x^2 + (y + 7)^2 = 25

Denote by C the centre of the circle. Let's say the coordinates of C are a and b.
We know that C is on the line x-y=7, so a-b=7 (*)

The slope of the line 4x+3y=4 is m=-4/3
the line passing through C and P is perpendicular on the line 4x+3y=4 (why?)
so its slope should be -1/m=3/4
But the slope of the line CP is (-4-b)/(4-a)
equate this to 3/4 and you get a relation of the form:
3a-4b=28 (**)
from (*) and (**) solve for a and b
a=0; b=-7
so you have the coordinates of the centre C(0;-7)
the radius of the circle is CP=5
the equation of the circle is:
x^2+(y+7)^2=25.

(x-h)² + (y-k)² = a², 2(x-h) + 2(y-k) = 0, y' = h-x / y-k
Hence, at (4, -4), y' = 4-h / 4+k = -4/3

If the center is on the line x-y = 7, h - k = 7, h = 7+k
Substitute back: 4-(7+k) / 4+k = -4/3
3(-3 - k) = -4(4 + k), k = -7, h = 0

a² = 4² + (-4+7)² = 25
Hence, x² + (y+7)² = 25