Determine whether or not the equation has a circle as its graph. x2 + 6x + y2 + 8y + 30 = 0?

Question

Determine whether or not the equation has a circle as its graph.

x2 + 6x + y2 + 8y + 30 = 0

Answer 1

No, write it in the form:
x^2 + 6x + 9 + y^2 + 8y + 16 + 5 = 0, or
(x + 3)^2 + (y + 4)^2 + 5 = 0, now you have a sum of 2 non-negative squares and a strictly positive 5, that sum cannot be 0. No real values of x and y satisfy that equation, the graph is void.

Answer 2

Circle is of the form w^2 + v^2 = r^2 where r is the radius of the circle

(x^2 + 6x) + 9 = (x + 3)^2
(y^2 + 8y) + 16 = (y + 4)^2

(x^2 + 6x) + (y^2 + 8y) + 30 = 0
(x + 3)^2 - 9 + (y + 4)^2 - 16 + 30 = 0
(x + 3)^2 + (y + 4)^2 = -5 and r = SQRT(5)i (imaginary)

It looks like it is not a circle (unless you are ok with circles that have an imaginary number for their radius)

Answer 3

Let's complete the square in x and y:
We get
x²+6x+9 + y²+8y+16+5 = 0.
So
(x+3)²+(y+4)² = -5.
Since the sum of 2 squares can't be negative,
the graph is not a circle. It's empty.

Answer 4

circle equations take the form (x – h)2 + (y – k)2 = r2

Completing the squares give:

x^2+6x+9+y^2+8y+16+5=0

(x+3)^2+(y+4)^2= -5

dealing with the real world, -5 cannot equal r^2

looks nice but no dice

Answer 5

The general equation for a circle is
(x-x0)^2+(y-y0)^2=R^2
in which (x0,y0), and R are the center coordinates and the circle radius.
The expression can be written as
(x+3)^2 -9 +(y+4)^2 -16 +30=0
(x+3)^2 +(y+4)^2 =-5
Since the right hand side is a negative number it is not a circle. In fact there is no possible real coordinate to satisfy this equation (or no curve) because right hand side should always be equal or greater than zero but left hand side is negative.