Question 256775: wHAT IS THE CENTER OF THE CIRCLE X RAISED TO THE 2ND POWER + Y RAISED TO THE SECOND POWER - 4X + 2Y -11 = 0
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! If I did this correctly, then the center of the circle should be (x,y) = (2,-1).
you wind up with the equation of:
(x-2)^2 + (y+1)^2 = 16
to graph this equation, you need to solve for y.
subtract (x-2)^2 from both sides of the equation to get:
(y+1)^2 = 16 - (x-2)^2
take square root of both sides of this equation to get:
y+1 = +/- sqrt(11-(x-2)^2)
subtract 1 from both sides of this equation to get:
y = -1 +/- sqrt(16-(x-2)^2)
graph this equation to get:
from the graph you can see that the center of the circle appears to be x,y = 2,-1.
if you freeze y at -1, then the horizontal range is from x = -2 to x = 6 with x = 2 right in the middle (+/- 4 each way).
if you freeze x at 2, then the vertical range is from y = -5 to y = 3 with y = -1 right in the middle (+/- 4 each way).
the center of the circle is definitely at x,y = (2,-1).
here's how it was derived:
your original equation is:
x^2 + y^2 - 4x + 2y - 11 = 0
reorder the terms to that the x's and the y's are together to get:
x^2 - 4x + y^2 + 2y - 11 = 0
complete the squares on x^2 - 4x to get (x-2)^2 - 4
complete the squares on y^2 + 2y to get (y+1)^2 - 1
your equation becomes:
(x-2)^2 - 4 + (y+1)^2 - 1 - 11 = 0
combine like terms to get:
(x-2)^2 + (y+1)^2 - 16 = 0
add 16 to both sides of the equation to get:
(x-2)^2 + (y+1)^2 = 16
you have just converted to the standard form of the equation for the circle.
that standard form is:
(x-h)^2 + (y-k)^2 = r^2
h is the x coordinate of the center of the circle.
k is the y coodinate of the center of the circle.
r is the radius of the circle.
the answer to problem is:
the center of the circle is at (x,y) = (2,-1).
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