How do you classify the conic #-3x^2-3y^2+6x+4y+1=0#?

1 Answer
Jan 10, 2017

Answer:

This is a circle with centre #(1, 2/3)# and radius #4/3#

Explanation:

Given:

#-3x^2-3y^2+6x+4y+1 = 0#

Looks like a circle: The coefficients of #x^2# and #y^2# are the same. Let's rearrange it a little...

Divide the equation by #-3# to get:

#x^2-2x+y^2-4/3y-1/3 = 0#

Complete the squares for #x# and #y#...

#x^2-2x+1+y^2-4/3y+4/9-16/9 = 0#

That is:

#(x-1)^2+(y-2/3)^2 = (4/3)^2#

This is in the form:

#(x-h)^2+(y-k)^2 = r^2#

which is the equation of a circle with centre #(h, k) = (1, 2/3)# and radius #r = 4/3#