Find the center of the circle with equation x2 + y2 - 10x + 6y + 27
x^2 + y^2 - 10x + 6y + 27 = 0 (x^2 - 10x + 25) + (y^2 + 6y + 9) = 7 (x - 5)^2 + (y + 3)^2 = 7 So the center of the circle is (5, -3)
Question
Updated 5/8/2014 12:43:12 AM
Rating
3
x^2 + y^2 - 10x + 6y + 27 = 0
(x^2 - 10x + 25) + (y^2 + 6y + 9) = 7
(x - 5)^2 + (y + 3)^2 = 7
So the center of the circle is (5, -3)
Confirmed by andrewpallarca [5/8/2014 12:48:35 AM]

Questions asked by the same visitor
Write an equation for the circle that satisfies the given set of conditions. endpoints of a diameter at (11, 18) and (-13, -19)
Question
Updated 5/8/2014 12:48:30 AM
The midpoint between (11, 18) and (-13, -19) is (-1, -1/2) which is the center of the circle.
The distance between (11, 18) and (-13, -19) is sqrt (1945) which is the diameter of the circle.
So, the equation of the circle is (x + 1)^2 + (y + 1/2)^2 = 1945/4
Confirmed by andrewpallarca [5/8/2014 12:52:33 AM]
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