Having a diameter w/ end points at (-3,4) and (9,8)

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perhaps you're referring to center-radius form?
yes po.
(x-3)² + (y-6)² = (2\sqrt{10})^2
(x-h)² + (y-k)² = r² is the center-radius form
pano po pag ito, Concetric w/ the circle described by the equation x2+y2-8y=0 and w/ radius of 5 units.

Answers

2014-11-27T22:36:47+08:00
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:
x_m =  \frac{x_2+x_1}{2}
x_m =  \frac{9+(-3)}{2}
x_m =  \frac{6}{2}
x_m = 3
for the y coordinate:
y_m =  \frac{y_2-y_1}{2}
y_m =  \frac{8+4}{2}
y_m =  \frac{12}{2}
y_m = 6
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The coordinates of the center is (3,6)
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To get the radius of the circle,
r =  \sqrt{(9-3)^2+(8-6)^2}
r =  2\sqrt{10}
r^2 = 40
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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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