Write the equation in standard form for a circle that has a diameter with endpoints (8,0) and (-8,0)

x²+y² = 64

Step-by-step explanation:

The standard form of writing the equation of a circle is expressed as;

(x-a)²+(y-b)² = r² where;

(a, b) is the centre of the circle

r is the radius of the circle

Given the diameter of the circle with endpoints (8,0) and (-8, 0)

d = √(x2-x1)²+(y2-y1)²

d = √(-8-8)²+(0-0)²

d = √(-16)²

d = √256

d = 16

radius r = 16/2

r = 8

The centre of the circle will be the midpoint of the coordinates

M = (8-8/2, 0+0/2)

M = (0/2, 0/2)

M = (0,0)

Hence the centre is at (0,0)

Get the required equation

Recall that (x-a)²+(y-b)² = r²

Substitute the centre and the radius

(x-0)²+(y-0)² = 8²

x²+y² = 64

This gives the required equation

is there answer choices?