Find intercepts of x^2+x+y^2+2y=0

Question

Is it a circle ?

Answer 1

x² + x + y² + 2y = 0

 

Complete the square to put the equation in the form (x-h)² + (y-k)² = r²

 

(x² + x + 1/4) + (y² + 2y + 1) = 0 + 1/4 + 1

 

(x + 1/2)² + (y + 1)² = 5/4 

 

[x - (-1/2)]² + [y - (-1)]² = (√5/2)²

 

This is a circle with center (-1/2, -1) and radius √5/2

Answer 2

x² + x + y² + 2y = 0

 

 

To find the x-intercepts, we set y=0 into the equation and solve for x.

 

x² + x = 0

 

x(x + 1) = 0

 

 

x = 0         and        x = -1

 

The x-intercepts are (0, 0) and (-1, 0).

 

 

To find the y-intercepts, we set x=0 into the equation and solve for y.

 

y² + 2y = 0

 

y(y + 2) = 0

 

y = 0         and        y = -2

 

The y-intercepts are (0, 0) and (0, -2).

 

 

The equation of a circle is    x² + y² = r²     when the centerpoint is located at the origin.  Since the given equation does not match this form, it is not a circle.