Graph the circle, having trouble finding the center and radius, I know I have to do something with completing the squares but I'm not sure how?
thanks
http://i48.tinypic.com/fngvx1.jpg
thanks
http://i48.tinypic.com/fngvx1.jpg
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x² + y² - 2x - 10y + 22 = 0
(x² - 2x) + (y² - 10y) = -22
(x² - 2x + 1) + (y² - 10y + 25) = -22 + 1 + 25
(x - 1)² + (y - 5)² = 4
The circle has radius 2 and is centered at (1,5)
(x² - 2x) + (y² - 10y) = -22
(x² - 2x + 1) + (y² - 10y + 25) = -22 + 1 + 25
(x - 1)² + (y - 5)² = 4
The circle has radius 2 and is centered at (1,5)
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Rearrange the equation so that the x terms are first, the y terms are next and the constant is on the right:
x² - 2x + y² - 10y = -22
Use the form (x - a )² = x² - 2ax + a² on the x terms. You do this by setting the middle term of the general form equal to the corresponding x term in the equation:
-2ax = -2x
ax = x
a = 1 and a² = 1
Add 1 to both sides:
x² - 2x + 1 + y² - 10x = -21
We just made the x terms a perfect square where a = 1:
(x - 1)² + y - 10y = -21
Now we use the same form to find a for the y terms:
-2ay = -10y
ay = 5y
a = 5 and a² = 25 so we add 25 to both sides:
(x - 1)² + (y - 5)² = 3
This is a circle with a center at (1, 5) and a radius of √3
x² - 2x + y² - 10y = -22
Use the form (x - a )² = x² - 2ax + a² on the x terms. You do this by setting the middle term of the general form equal to the corresponding x term in the equation:
-2ax = -2x
ax = x
a = 1 and a² = 1
Add 1 to both sides:
x² - 2x + 1 + y² - 10x = -21
We just made the x terms a perfect square where a = 1:
(x - 1)² + y - 10y = -21
Now we use the same form to find a for the y terms:
-2ay = -10y
ay = 5y
a = 5 and a² = 25 so we add 25 to both sides:
(x - 1)² + (y - 5)² = 3
This is a circle with a center at (1, 5) and a radius of √3