SOLUTION: wHAT IS THE CENTER OF THE CIRCLE X RAISED TO THE 2ND POWER + Y RAISED TO THE SECOND POWER

SOLUTION: wHAT IS THE CENTER OF THE CIRCLE X RAISED TO THE 2ND POWER + Y RAISED TO THE SECOND POWER - 4X + 2Y -11 = 0

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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