A ridiculously hard circles question?

Question

Find the centre and radius of the circle x^2+y^2-2x+8y-3=0

Answer 1

Complete the square.

(x^2 - 2x + 1) + (y^2 + 8y + 16) = 3 + 1 + 16

(x - 1)^2 + (y + 4)^2 = 20

Now you tell me what you're looking for.

Answer 2

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where h and k are the x- and y-coordinates of the center and r is the radius.

You have

x²+y²-2x+8y-3=0

Start factoring by completing the squares...

x²-2x+y²+8y=3
x²-2x +1+y²+8y+16 =3+1+16

(x-1)² + (y+4)² = 20

The center of your circle is at x = 1, y = -4 and the radius is sqrt(20).

Answer 3

complete the square

(x^2 -2x + ?) + (y^2 +8y + ?) = 3

[1/2(-2) ]^2 = 1 this adds into x group and to right side
[1/2(8)]^2 = 16 this adds into y group and to right side

(x^2 - 2x + 1) + (y^2 +8x + 16 ) = 3+1+16
factor
(x -1)^2 + (y + 4)^2 = 20

Center (1, -4) and radius = sqrt(20)