SOLUTION: Find the center of a circle with the equation:x^2+y^2-6x + 10y +24 = 0

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Question 279776: Find the center of a circle with the equation:x^2+y^2-6x + 10y +24 = 0




Found 2 solutions by Fombitz, scott8148:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square in x and y.
+x%5E2%2By%5E2-6x+%2B+10y+%2B24+=+0+
x%5E2-6x%2By%5E2%2B+10y+%2B24+=+0
%28x%5E2-6x%2B9%29%2B%28y%5E2%2B+10y+%2B25%29%2B24+=+9%2B25+
%28x-3%29%5E2%2B%28y%2B5%29%5E2%2B24+=+34+
%28x-3%29%5E2%2B%28y%2B5%29%5E2+=+10+
The general equation for a circle centered at (h,k) with a radius R is:
%28x-h%29%5E2%2B%28y-k%29%5E2=R%5E2
This circle is centered at (3,-5).

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + y^2 - 6x + 10y = -24

x^2 - 6x + 9 + y^2 + 10y + 25 = 10

(x - 3)^2 + (y + 5)^2 = 10

center is (3, -5)