Graph the circle
Given x^2+ y^2- 4x + 8y + 16 = 0
When solving for a circle I did the following steps:
Place in standard notation.
x^2- 4x + y^2+ 8y = -16
Complete the square.
x^2- 4x + 4 + y^2+ 8y + 16 = -16 + 4 +16
(x - 2) ^2 + (y + 4) ^2= 4 = 2^2
The equation is now written in standard form.
The center is (2 ,-4) and the radius is 2.

Question

Graph the circle
Given x^2+ y^2- 4x + 8y + 16 = 0 
When solving for a circle I did the following steps:
Place in standard notation. 
x^2- 4x + y^2+ 8y = -16 
Complete the square.
x^2- 4x + 4 + y^2+ 8y + 16 = -16 + 4 +16
(x - 2) ^2 + (y + 4) ^2= 4 = 2^2
The equation is now written in standard form. 
The center is (2 ,-4) and the radius is 2.

anonymous · Answer

So my circle would have points that look like this:

Center= 2,-4
2,-2
0,-4
4,-4
2,-6

anonymous · Answer

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