Prove that AB-diameter of circle (x-2) ^ 2 + (y-1) ^ 2 = 10 if A (5; 2), B (-1; 0)

Knowing the coordinates of points A and B, we determine the length of the diameter AB.

AB = √ ((X2 – X1) ^ 2 + (Y2 – Y1) ^ 2) = √ ((- 1 – 5) ^ 2 + (0 – 2) ^ 2) = √ (36 + 4) = √40 = 2 * √10 cm.

Then the radius of the circle is: AO = AB / 2 = 2 * √10 / 2 = √10 cm.

Determine the coordinates of the point O, the center of the circle.

X0 = (X1 + X2) / 2 = (5 – 1) / 2 = 2.

Y0 = (Y1 + Y2) / 2 = (2 – 0) / 2 = 1.

O (2; 1).

(X – X0) ^ 2 + (Y – Y0) ^ 2 = R ^ 2.

(X – 2) ^ 2 + (Y – 1) ^ 2 = (√10) ^ 2 = 10.

Since the equations are the same, AB is the diameter of the circle, which was required to be proved.