Mathematics: If the endpoints of the diameter of a circle are (6, -2)and (-6 ,2)what is the standard form equation of the circle

its 8 or 3Step-by-step explanation:it could be both i hope this helps...

If the endpoints of the diameter of a circle are

kelli151

02.04.2021

x^2 + y^2 - 40 = 0

Step-by-step explanation:

Given

(h_1,k_1) = (6,-2)

(h_2,k_2) = (-6,2)

Required

Determine the equation of the circle

First, calculate the midpoint of the given coordinates:

$(h,k) = \frac{1}{2}(h_1+h_2,k_1+k_2)$

$(h,k) = \frac{1}{2}(6-6,-2+2)$

$(h,k) = \frac{1}{2}(0,0)$

(h,k) = (0,0)

Next, calculate the radius.

This is the distance between (0,0) calculated above and any of (6,-2) (-6,2)

Using:

(0,0) and (-6,2) , the radius is:

$r = \sqrt{(0 - (-6))^2 + (0-2)^2}$

$r = \sqrt{40}$

The equation is then calculated using:

(x - h)^2 + (y - k)^2 = r^2

Where:

$r = \sqrt{40}$ and (h,k) = (0,0)

$(x - 0)^2 + (y - 0)^2 = (\sqrt{40})^2$

(x )^2 + (y)^2 = 40

x^2 + y^2 = 40

x^2 + y^2 - 40 = 0

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