Question 429551
Since the equation includes both an {{{x^2}}} term and a {{{y^2}}} term, it looks like it would be the equation of a conic section.
{{{x^2+y^2-4x+12y-6 = 0}}} Start by grouping the x- and y-terms.
{{{(x^2-4x)+(y^2+12y)-6 = 0}}} Next add 6 to both sides.
{{{(x^2-4x)+(y^2+12y) = 6}}} Now complete the square in both x and y.
{{{(x^2-4x+4)+(y^2+12y+36) = 6+4+36}}} Simplify.
{{{(x-2)^2+(y+6)^2 = 46}}} Compare this with the general form of the equation of a circle with center at (h, k) and radius r.
{{{(x-h)^2+(y-k)^2 = r^2}}}
So what you have is the equation of a circle with its center at (2, -6) and a radius {{{r = sqrt(46)}}}.