When graphed, the circle with equation
x2 + y2 - 14x + 10y + 65 = 0
will lie entirely in quadrant

Recall what each quadrant means.
Quadrant I (x, y)
Quadrant II (-x, y)
Quadrant III (-x, -y)
Quadrant IV (x, -y)

Now, we just need to complete the square twice to produce the general form of a circle.

x^2 - 14x + 49 + y^2 + 10y + 25 + 65 - 49 - 25 = 0
(x - 7)^2 + (y + 5)^2 - 9 = 0
(x - 7)^2 + (y + 5)^2 = 9

Centre at (7, -5)
Always lies in Quadrant IV

I’m pretty sure the slope would be 1 since the x and y values increase by 1