Question

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

Answer 1

The formula for a circle is:

\(\large\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}\)

\(\bf{(h,k)}\) is your vertex.

\(\bf{r}\) is your radius.

So determine from there.

Answer 2

x2 + y2 + y – 1 = 0

Answer 3

@dude @Vocaloid @TheSmartOne @563blackghost

Answer 4

*general form* is written as \(f(x,y) = 0\)

so multiply/ie FOIL it all out and set LHS to zero

Answer 5

@Shadow is my answer correct?

Answer 6

\[x^2 + (y + 1)^2 = 2\]
\[x^2 + (y + 1)(y + 1) = 2\]
\[x^2 + y^2 + 2y + 1 = 2\]
\[x^2 + y^2 + 2y - 1 = 0\]

Answer 7

\[1 \times y = y\]
You do this 2x, and them add them
\[y + y = 2y\]

Answer 8

so, the answer is x^2+y^2+2y-1=0?

Answer 9

Yes. Do you see how I got there?

Answer 10

yes, thanks for explaining it so well

Answer 11

though, how do you get 2y, in the third step? is it blc there are two Ys?

Answer 12

@Shadow ?

Answer 13

FOIL means First, Outer, Inner, and Last

We have:
\[(y + 1)^2 \]
This is also represented as:
\[(y + 1)(y + 1)\]
We multiply the first:
\[y \times y = y^2\]
We multiply the outer:
\[y \times 1 = y\]
We multiply the inner:
\[1 \times y = y\]
We multiply the last:
\[1 \times 1 = 1\]
Sum them, or add them up:
\[y^2 + y + y + 1 = y^2 + 2y + 1\]