MATH SOLVE

4 months ago

Q:
# the endpoints of the diameter of a circle are (β6, 6) and (6, β2), what is the standard form equation of the circle?

Accepted Solution

A:

The equation of a circle in standard form is

[tex] (x - h)^2 + (y - k)^2 = r^2 [/tex]

where (h, k) is the center of the circle, and r is the radius if the circle.

We need to find the radius and center of the circle.

We are given a diameter, so to find the center, we need the midpoint of the diameter.

M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)

The center is (0, 2).

To find the radius, we find the length of the given diameter and divided by 2.

[tex] d = \sqrt{(-6 - 6)^2 + (6 - (-2))^2)} [/tex]

[tex] d = \sqrt{144 + 64} [/tex]

[tex] d = \sqrt{208} [/tex]

[tex] r = \dfrac{d}{2} = \dfrac{\sqrt{208}}{2} = \dfrac{\sqrt{208}}{\sqrt{4}} = \sqrt{52} [/tex]

[tex] (x - 0)^2 + (y - 2)^2 = (\sqrt{52})^2 [/tex]

[tex] x^2 + (y - 2)^2 = 52 [/tex]

[tex] (x - h)^2 + (y - k)^2 = r^2 [/tex]

where (h, k) is the center of the circle, and r is the radius if the circle.

We need to find the radius and center of the circle.

We are given a diameter, so to find the center, we need the midpoint of the diameter.

M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)

The center is (0, 2).

To find the radius, we find the length of the given diameter and divided by 2.

[tex] d = \sqrt{(-6 - 6)^2 + (6 - (-2))^2)} [/tex]

[tex] d = \sqrt{144 + 64} [/tex]

[tex] d = \sqrt{208} [/tex]

[tex] r = \dfrac{d}{2} = \dfrac{\sqrt{208}}{2} = \dfrac{\sqrt{208}}{\sqrt{4}} = \sqrt{52} [/tex]

[tex] (x - 0)^2 + (y - 2)^2 = (\sqrt{52})^2 [/tex]

[tex] x^2 + (y - 2)^2 = 52 [/tex]