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Question 152653: Complete the square and write the equation in standard form. Then give the center and radius of the circle.
x2 + y2 - 2x - 4y - 4 = 0
: Complete the square and write the equation in standard form. Then give the center and radius of the circle.
x2 + y2 - 2x - 4y - 4 = 0

Answer by mducky2(55) About Me  (Show Source):
You can put this solution on YOUR website!
It's easier to start by separating the x's from the y's:
x^2 + y^2 - 2x - 4y - 4 = 0
(x^2 - 2x) + (y^2 - 4y) - 4 = 0

In order to complete the square, we can use this equation:
x^2 + bx + (b/2)^2 = (x + b/2)^2

Now let's make the x's fit into this equation:
x^2 - 2x
We know that b = -2. But make sure to subtract whatever you add on so that everything adds up back in the original formula:
x^2 - 2x + 1^2 - 1^2
x^2 - 2x + 1 - 1
(x + 1)^2 - 1

Now let's make the y's fit into this equation:
y^2 - 4y
We know that b = -4
y^2 - 4y + 2^2 - 2^2
y^2 - 4y + 4 - 4
(y - 4)^2 - 4

Putting everything back into the original equation, we can now write it in standard form:
(x^2 - 2x) + (y^2 - 4y) - 4 = 0
((x + 1)^2 - 1) + ((y - 4)^2 - 4) - 4 = 0

Moving all the excess numbers to the side:
(x + 1)^2 + (y - 4)^2 - 4 - 1 - 4 = 0
(x + 1)^2 + (y - 4)^2 - 9 = 0
(x + 1)^2 + (y - 4)^2 = 9

The center of a circle (h,k) and the radius r can be found using this equation:
(x - h)^2 + (y - k)^2 = r^2

Back to the equation:
(x + 1)^2 + (y - 4)^2 = 9
It looks like h = -1 and k = 4. Therefore the center of the circle is (-1,4).

It also looks like r2 = 9. Let's solve for r:
r^2 = 9
r = 3
Therefore, the radius of the circle is 3.