Question

find an equation of the form (3) for the circle with the given equation: 2x^2+2y^2+2x+2y-1=0

Answer 1

form (3)?

Answer 2

the form 3 is just (x-h)^2+(y-k)^2=r^2

Answer 3

oh ok
first divide by 2
x^2 + y^2 + x + y = 0.5
(x + 0.5)^2 - 0.25 + (y + 0.5)^2 - 0.25 = 0.5
(x + 0.5)^2 + (y + 0.5)^2 = 1

Answer 4

find the radius and the center point

Answer 5

that last equation that @joeywhite wrote should give you the center and radius because it is what's called center-radius form.

Answer 6

i know that the radius is 1^2 which is 1 but what is the center?

Answer 7

compare the equation to the general form

(x-h)^2+(y-k)^2=r^2

(x + 0.5)^2 + (y + 0.5)^2 = 1

the center is (h, k) and r is radius

so for example the x coordinate of center is
-h = + 0.5
h = -0.5