Question

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

Answer 1

Answer:

$x^2 + y^2 - 40 = 0$

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 $(h,k) = (0,0)$

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

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

$x^2 + y^2 = 40$

$x^2 + y^2 - 40 = 0$