If the endpoints of the diameter of a circle are (8, 6) and (2,0), what is the standard form equation of the circle?
A)
(x + 5)2 + (y + 3)2 = 18
(x + 5)2 + (y + 3)2 = 3.72
(x - 5)2 + (y - 3)2 = 18
D)
(x - 5)2 + (y - 3)2 = 32

The centre of the circle is at the midpoint of The diameter = (8+2)/2, (6+0)/2 = (5,3) so the left side of the equation is (x-5)^2+(Y-3)^2. Now we need to work out the radius:

The diameter = square root(8-2)^2+6^2) = square root of 72 = 6sqrt2.

So the radius = 3sqrt2 and r^2=18.

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If the endpoints of the diameter of a circle are (8, 6) and (2,0), what is the standard form equation of the circle? A) (x + 5)2 + (y + 3)2 = 18 (x + 5)2 + (y + 3)2 = 3.72 (x - 5)2 + (y - 3)2 = 18 D) (x - 5)2 + (y - 3)2 = 32

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If the endpoints of the diameter of a circle are (8, 6) and (2,0), what is the standard form equation of the circle?
A)
(x + 5)2 + (y + 3)2 = 18
(x + 5)2 + (y + 3)2 = 3.72
(x - 5)2 + (y - 3)2 = 18
D)
(x - 5)2 + (y - 3)2 = 32