Find the center and radius of a circle with this equation.

Okay, I don't need the answer to this problem since I already have it (it's in the back of the book). I just want to know HOW to solve this problem. Okay, here's the equation...

x^2 + y^2 - 2x +4y - 4 = 0

x² + y² – 2x + 4y – 4 = 0 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ group variables
x² – 2x + ... + y² + 4y + ... = 4 ∙ ∙ ∙ ∙ ∙ ∙ ∙ complete squares
x² – 2x + 1 + y² + 4y + 4 = 4 + 1 + 4 ∙ ∙ ∙rewrite
(x – 1)² + (y + 2)² = 9

center at (1, -2), radius 3

1) Grouping the given one it is: (x² - 2x) + (y² + 4y) - 4 = 0

2) Make every part as perfect square by adding b²/4a [b stands for coefficient of (x,y) and a stands for coefficient of (x², y²) respectively]
Then to balance subtract the same.

==> The equation changes as:
(x² - 2x + 1) + (y² + 4y + 4) - 4 - (1+4) = 0

==> (x - 1)² + (y + 2)² = 9 = 3²

Wish you can now complete.

3) Alternatively, you may apply:

If a circle equation is given by x² + y² + 2gx + 2fy + c = 0,
then its centre is (-g, -f) and radius = √(g² + f² - c)

Follow these steps to solve this problem:

1. Rearrange: x^2-2x+y^2+4y=4
2. Complete the squares for both x and y (you should know this by now):

x^2 -2x + y^2 + 4y = 4
=>(x^2-2x+1-1)+(y^2 + 4y +4-4) = 4
=> (x-1)^2 - 1 + (y+2)^2 - 4 = 4
=> (x-1)^2 + (y+2)^2 = 9

The centre point is always (h,k) and radius r where: (x-h)^2 + (y-k)^2 = r^2

Therefore, the centre point is one - PLUS one for x (since its x-h and y-k, not x+h or y+k) and MINUS two, -2, for y.

The centre point is (1, -2). Radius = r, and if r^2 = 9, then r = 3.

Therefore, the centre point is (1,-2), and r = 3.

Cheers,

SC!