What is the equation of a circle whose diameter's end points are at 2 -3 and 8 7?

The mid point of the diameter is the circle's centre:

Mid point of (2, -3) and (8, 7) is ((2 + 8)/2, (-3 + 7)/2) = (5, 2)

The radius is half the diameter, but as the equation for a circle requires the radius squared:

radius = diameter/2

→ radius² = diameter² / 4

Diameter (and diameter²) can be found using Pythagoras:

diameter² = change_in_x² + change_in_y²

= (8 - 2)² + (7 - -3)²

= 6² + 10²

= 136

→ radius² = 136/4 = 34

Circle with centre (X, Y) and radius r has equation:

(x - X)² + (y - Y)² = r²

→ (x - 5)² + (y - 2)² = 34

→ x² -10x + 25 + y² - 4y + 4 = 34

→ x² + y² - 10x - 4x - 5 = 0