Note that the diameter will surely pass through the center. And the the center is located at the midpoint of the diameter.
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The coordinates of the center is just the midpoint of the line formed by connecting the two endpoints of the diameter.
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Use midpoint formula
for the x coordinate:
[tex]x_m = \frac{x_2+x_1}{2} [/tex]
[tex]x_m = \frac{9+(-3)}{2} [/tex]
[tex]x_m = \frac{6}{2} [/tex]
[tex]x_m = 3[/tex]
for the y coordinate:
[tex]y_m = \frac{y_2-y_1}{2} [/tex]
[tex]y_m = \frac{8+4}{2} [/tex]
[tex]y_m = \frac{12}{2} [/tex]
[tex]y_m = 6[/tex]
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The coordinates of the center is (3,6)
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To get the radius of the circle,
[tex]r = \sqrt{(9-3)^2+(8-6)^2} [/tex]
[tex]r = 2\sqrt{10} [/tex]
[tex]r^2 = 40[/tex]
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Equation for the circle:
(x-h)² + (y-k)² = r²
where (h,k) is the center r is the radius
(x-3)² + (y-6)² = 40
x² - 6x + 9 + y² - 12y + 36 = 40
x² + y² - 6x - 12y + 36 + 9 - 40 = 0
x² + y² - 6x - 12y + 5 = 0
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