Question

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

Answer 1

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

Explanation:

From the given equation `x^2+y^2+2x-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+2x-3=0`
`x^2+2x+y^2=3`
add 1 to both sides of the equation
`x^2+2x+1+y^2=3+1`
`(x^2+2x+1)+y^2=4`
`(x+1)^2+(y-0)^2=2^2`
it takes the form
`(x-h)^2+(y-k)^2=r^2`
with center at `(-1, 0)` with radius `r=2`

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