Given: x^2+y^2+4x-2y+1=0, write in standard form for a circle.?

Answer 1

you take the coefficients of x and y which are 4 and -2 respectively. you divide each by 2 and then square the answer:
(4/2)^2 = 4
(-2/2)^2 = 1
you add to the equation +4 -4, and +1 -1
This sounds like nonsence, but you'll see that:

x^2 +4x (+4 -4) + y^2 -2y (+1 -1) +1 = 0
x^2 +4x +4 + y^2 -2y +1 -4 -1 +1 =0
(x+2)^2 + (y-1)^2 -4 = 0
(x+2)^2 + (y-1)^2 = 4

the equation of the circle is (x-a)^2 + (y-b)^2 = r^2
a is the x-coordinate and b is the y-coordinate of the centre of the circle, r is the radius.

the centre is ( -2, 1) and the radius is √4 =2

Answer 2

Standard form: (x - h)^2 + (y - k)^2 = r^2 where (h,k) is the center and r is the radius.

1) First group terms. get all numbers on the right side, and all y's together and all x's together on the left:

x^2 + y^2 + 4x - 2y + 1 = 0
(x^2 + 4x) + (y^2 - 2y) = -1

2) Now, complete the square for each of (x^2 + 4x) and (y^2 - 2y)

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

You add 4 and 1 to the right side to keep the equation balanced. It's a rule of algebra. You add them to the left (completing the squares) to get a trinomial that is a perfect square, i.e. the same two binomials multiply to get that trinomial

(x + 2)^2 + (y - 1)^2 = 4
(-2, 1) is the center and 2 is the radius.

Answer 3

A circle needs a center and a radius.
The standard form is (x-x1)^2+(y-y1)^2= r^2, where (x1,y1) is the center coordinates, while r is the radius.
x^2+4x - we know these terms, to square it as the standard form demands - add 4.
y^2-2y needs a 1.
So: x^2+y^2+4x- 2y +1+4-4+1-1=0
So: (x+2)^2+(y-1)^2-4=0
So (x+2)^2+(y-1)^2=4 there.

Answer 4

x^2 + y^2 + 4x - 2y + 1 = 0
x^2 + 4x + y^2 - 2y + 1 = 0
(x^2 + 4x) + (y^2 - 2y) + 1 = 0
(x^2 + 4x + 4 - 4) + (y^2 - 2y + 1 - 1) + 1 = 0
((x + 2)^2 - 4) + ((y - 1)^2 - 1) + 1 = 0
(x + 2)^2 - 4 + (y - 1)^2 - 1 + 1 = 0
(x + 2)^2 + (y - 1)^2 - 4 = 0
(x + 2)^2 + (y - 1)^2 = 4

ANS : (x + 2)^2 + (y - 1)^2 = 4

Answer 5

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