How do you classify #13x² + 13y² - 26x + 52y = -78#?

1 Answer
Oct 30, 2015

Answer:

It has the form of a circumference, but it would have negative radius, so it represents no point in the plane.

Explanation:

First of all, note that we have, coefficients apart, two quadratic terms (#x^2# and #y^2#), and two linear terms (#x# and #y#).

Also, we see that the coefficients of the quadratic terms are equal (they're both #13#). This means that we can divide the whole expression by #13#, obtaining

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

Now we can complete the squares:

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

Which we can rewrite into

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

We know that a circumference with center #(x_0,y_0)# and radius #r# is represented by the equation

#(x-x_0)^2 + (y-y_0)^2 = r^2#.

Our equation mimics this one, except for the fact that we have #r^2=-1#, which is obviously impossible using real numbers. In this case, the set of points satisfying the equation is the empty set.