How do you graph x^2 + y^2 + 2x - 3 = 0?

it is a circle with radius $r = 2$ and center at $\left(h , k\right) = \left(- 1 , 0\right)$

Explanation:

From the given equation ${x}^{2} + {y}^{2} + 2 x - 3 = 0$
perform completing the square method to determine if its a circle, ellipse, hyperbola. There are 2 second degree terms so we are sure it is not parabola

${x}^{2} + {y}^{2} + 2 x - 3 = 0$
${x}^{2} + 2 x + {y}^{2} = 3$
add 1 to both sides of the equation
${x}^{2} + 2 x + 1 + {y}^{2} = 3 + 1$
$\left({x}^{2} + 2 x + 1\right) + {y}^{2} = 4$
${\left(x + 1\right)}^{2} + {\left(y - 0\right)}^{2} = {2}^{2}$
it takes the form
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
with center at $\left(- 1 , 0\right)$ with radius $r = 2$

See the graph of ${x}^{2} + {y}^{2} + 2 x - 3 = 0$
graph{(x+1)^2+(y-0)^2=2^2[-10,10,-5,5]}
God bless .... I hope the explanation is useful.