Can you get this equation "2x^+2y^2+4x-8y-22" into a standard circle equation form?

Question

The standard circle equation for is (x-h)^2+(y-k)^2= r^2

Answer 1

x² + y² + 2x - 4y - 11 = 0
(x² + 2x) + (y² - 4y) = 11
(x² + 2x + 1) + (y² - 4y + 4) = 11 + 1 + 4
(x + 1)² + (y - 2)² = 16
[ x - (-1) ] ² + [ y - 2 ] ² = 4 ²
centre (- 1 , 2)
radius 4

Answer 2

you are able to comprehensive the sq. for the two x and y 4x^2 - 8x ....+y^2 + 3y ..... = 19 4(x^2 - 2x ......) + y^2 + 3y .......=19 discover the selection mandatory = (b/2)^2 FOR x want +a million for y want +9/4... stability equation while putting in blamks 4(x^2-2x+ a million) + y^2 + 3y + 9/4 = 19 ( upload to the two facets)+4+9/4 4(x-a million)^2 + (y+3/2)^2 = 19+4+(9/4) that is not A CIRCLE!!!!! that is an ellipse! Divide by capacity of four (x-a million)^2 + (y+3/2)^2 / 4 = one 0 one/sixteen OR ordinary form for ellipse: (x-a million)^2 + (y+3/2)^2 = one 0 one ____________________ . . . . . . . . . 2^2..........4^2

Answer 3

Yes, complete the square. I'll assume that your equation is in standard form (equals zero)

First, because all coefficients are even, divide by 2 to make things easier and get:
x^2 + y^2 + 2x - 4y - 11 = 0

Re-order the terms:
x^2 + 2x + y^2 - 4y - 11 = 0

Complete the squares by adding and subtracting the missing numbers:
x^2 + 2x + 1 - 1 + y^2 - 4y + 4 - 4 - 11 = 0
(x + 1)^2 - 1 + (y - 2)^2 - 4 - 11 = 0

Simplify:
(x + 1)^2 + (y - 2)^2 - 16 = 0
(x + 1)^2 + (y - 2)^2 = 16
(x + 1)^2 + (y - 2)^2 = 4^2

So this circle has radius 4 and is centered at (-1,2).