To find the equation for a circle that has the center at $(-1 , 4)$ and passes through the point $(3, -2)$, the right hand side of your equation is fine: you used the center of the circle correctly. So we have: $(x+1)^2+(y-4)^2 = r^2$
To determine the radius of the circle, you need to compute the distance between the given point on the circumference of the circle and the center of the circle:
$r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(3 - (-1))^2 + (-2 - 4)^2} = \sqrt{4^2+ (-6)^2} =\sqrt{52}$
Since we need to square $r$, we can use $r^2 = 52$.
$(x+1)^2+(y-4)^2 = 52$
(I simply chose $(3, -2) = (x_1, y_1)$ and $(-1, 4) = (x_2, y_2)$, but you get the same result if the points were assigned differently. Since the difference between the $x$-coordinates and the difference between the $y$-coordinates are both squared, which points you take as point 1 and point 2 is of no concern.)