Is the equation #13x² + 13y² - 26x + 52y = -78 # a line, parabola, ellipse, hyperbola, or circle?

2 Answers
Jul 25, 2018

Answer:

No - it contains no real points.

Explanation:

Given:

#13x^2+13y^2-26x+52y=-78#

First note that all of the coefficients are divisible by #13#, so let's divide the whole equation by #13# to get:

#x^2+y^2-2x+4y=-6#

Adding #1+4=5# to both sides of the equation and rearranging slightly, this becomes:

#x^2-2x+1+y^2+4y+4 = -1#

That is:

#(x-1)^2+(y+2)^2 = -1#

Note that this is always false for real values of #x# and #y#, so this equation describes an empty set of values in the #x# #y# plane.

Answer:

Imaginary circle

Explanation:

The given equation:

#13x^2+13y^2-26x+52y=-78#

#13(x^2+y^2-2x+4y)=-78#

#x^2+y^2-2x+4y=-6#

#(x^2-2x+1)+(y^2+4y+4)-5=-6#

#(x-1)^2+(y+2)^2=-1#

#(x-1)^2+(y+2)^2=i^2#

Above equation shows a circle with a radius #r=i# i.e. an imaginary circle with unit radius.