Question

How do you graph `2x^2+8x+2y^2=0`?

Answer 1

See the accompanying graph
graph{2x^2+8x+2y^2=0 [-5, 5, -2.5, 2.5]}

Explanation:

Equation of circle in Standard form is
`(x-h)^2+(y-k)^2=r^2`
Here `(h,k)` are the coordinates of the centre of the circle and `r` is its radius.

Given equation is
`2x^2+8x+2y^2=0`

We see that 2 can be factored out both sides of the equation. Thus we obtain

`x^2+4x+y^2=0`

This can be written in the standard form by adding 4 to the both sides and by using the identity `(a+b)^2=a^2+2ab+b^2`.

`x^2+4x+4+y^2=0+4`

`(x+2)^2+(y-0)^2=2^2`
or `(x-[-2])^2+(y-0)^2=2^2`
This gives us `(-2,0)` coordinates of the centre. Also `r=2 " units"`
We also observe that origin `(0,0)` lies on the circle.