Question 1078261: The equation of a circle is x^2+y^2-4x+2y-11=0. What are the center and the radius of the circle? Show work please because I don't understand this on my homework and I need to try to figure it out!
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Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x^2+y^2-4x+2y-11=0.
Separate the x s and the y s, move the 11 to the other side, complete the square, which is taking 1/2 th y value and squaring it, adding it to both sides of the equation.
x^2-4x+ y^2-2y= 11
to complete the square for the x, take half of 4 (2), square it (4), and add it to both sides.
x^2+4x+4+y^2-2y=11+4
Now do the same for the y, half of -2 (-1), square it (1), and add it to both sides.
x^2+4x+4+y^2-2y+1=11+4+1
The left side is two perfect squares, (x+2)^2+(y-1)^2=16, which is 11+4+1. This is the equation of a circle
The center: the x-coordinate is -2, take the number and change the sign. The y-coordinate would then be +1.
The radius is the square root of 16, or 4.
The center is at (-2, 1), with radius 4.
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