Transpose $x^2 + y^2 + 2x + 2y + 1 = 0$ to find $y$ in terms of $x$

Question

I'm self-studying from Stroud & Booth's phenomenal "Engineering Mathematics", and am stuck on a problem from the end of the fourth chapter, "Graphs".

Namely, I'm to transpose the following equation to find $y$ in terms of $x$:

$$x^2 + y^2 + 2x + 2y + 1 = 0$$

I'm just completely stumped as to how to solve this, as there is the $y^2 + 2y$ component.

I've gotten as far as:

$$ y^2 + 2y = -x^2 - 2x - 1$$

but have no idea as to how to proceed...

Answer 1

It is $$y^2+2y+(x+1)^2=0$$ and using the quadratic formula we obtain $$y_{1,2}=-1\pm\sqrt{1-(x+1)^2}$$ It is $$ax^2+bx+c=0$$ and the formula is given by $$x_{1,2}=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$

Answer 2

Try completing the square i.e. $$ y^2+2y=(y+1)^2 -1 $$

Similarly for $x$.

Answer 3

Hint:

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