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

1 Answer
Feb 13, 2016

Answer:

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.