Find the equation of a circle that has a center (-3,1) and passes through (4,-4)?

r² = [ - 4 - 1 ]² + [ 4 + 3 ]²
r² = 25 + 49
r² = 74

[ x + 3 ]² + [ y - 1 ]² = 74

First find the radius by using Pythagoras.
r^2 = (4 - - 3)^2 + (-4 - 1)^2
r^2 = 7^2 + 5^2
r^2 = 49 + 25 = 74
r = sqrt(74)
Circle eq'n
( x - - 3)^2 + ( y - 1)^2 = 74
x^2 + 6x + 9 + y^2 - 2y + 1 = 74
x^2 + y^2 + 6x - 2y = 64

radius of the circle, r = √[ (4 - (-3))^2 + (- 4 - 1)^2

= √ 49 + 25

= √74

Equation of a circle of centre (a, b) and radius r is given by

( x - a)^2 + (y - b)^2 = r^2

Equation of required circle

(x + 3)^2 + (y - 1)^2 = 74

The general formula for a circle with center (h, k) and radius r is:
(x - h)² + (y - k)² = r²

In your case, you know the center but not the radius. But the radius is the distance from the center to a point on the circle. So let's use the distance between those two points:

Difference of x-coordinates --> 4 - (-3) = 7
Difference of y-coordinates --> 1 - (-4) = 5

7² + 5² = r²
49 + 25 = r²
74 = r²

Now you have what you need.
Center (-3, 1) --> h = -3, k = 1
Radius --> r² = 74

Answer:
(x + 3)² + (y - 1)² = 74