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

1 Answer

Answer:

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.