Question

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.
x2 + y2 − 2x + 2y − 1 = 0
x2 + y2 − 4x + 4y − 10 = 0
x2 + y2 − 8x − 6y − 20 = 0
4x2 + 4y2 + 16x + 24y − 40 = 0
5x2 + 5y2 − 20x + 30y + 40 = 0
2x2 + 2y2 − 28x − 32y − 8 = 0
x2 + y2 + 12x − 2y − 9 = 0

Answer 1

Wow. You're encircled by circles!! Looks like you'll have to go through all of the equations and determine the radii, then rearrange the circles in ascending order by radius length.

First problem: x2 + y2 − 2x + 2y − 1 = 0
Note: Please use x^2 (not x2) to denote "the square of x."

Rewriting this by grouping the x-terms and the y-terms:\[x^2 -2x +y^2 +2y = 1\]

Answer 2

We have to "complete the square for both the x^2 -2x and the y^2+2y groups.
Know how to do this?
Starting with x^2-2x, take half of the coefficinet of x and square it.
In other words, take half of -2, which gives us -1, and square -1, to get +1.

Add, then subtract, 1 from x^2-2x: x^2-2x+1-1.
Perform the same operations on the y terms: y^2+2y+1-1

Rewrite all this as \[x^2-2x+1 + y^2 +2y + 1 -1 -1 =1, \]

Answer 3

... or ...\[(x-1)^2+(y+1)^2 =3 = (\sqrt{3})^2\]

Answer 4

The radius of this circle is 2. The center of the circle is at (1, -1).

Please try the next equation. If you need more info and/or practice on how to "complete the square," please ask.

Answer 5

thanks. i understand how to do it now @mathmale

Answer 6

I'm really happy for you. Good luck!