Question

Determine the general form of the equation for the circle (x – 1)^2 + (y + 2)^2 = 3.

Answer 1

@Vocaloid pls help

Answer 2

x2 + y2 – x + y + 2 = 0
@Vocaloid ?

Answer 3

I think you messed up on your expansion somewhere, remember your binomial expansion rules

(a+b)^2 = a^2 + 2ab + b^2

therefore
(x – 1)^2 + (y + 2)^2 - 3 = 0
becomes

(x^2 - 2x + 1) + (y^2 + 4y + 4) - 3 = 0, combine like terms

Answer 4

still there? i've done the expansion you just need to combine like terms (the integers) and then just re-arrange in general form

Answer 5

x2 + y2 – x + y + 3 = 0
@Vocaloid ?

Answer 6

(x^2 - 2x + 1) + (y^2 + 4y + 4) - 3

x^2 + y^2 - 2x + 4y + 2 = 0
not sure how you're getting -x and + y