Question 61071
Find the centre and radius of the circle:
{{{x^2 + y^2 - 12x - 16y - 21 = 0}}} First, add 21 to both sides.
{{{x^2 + y^2 - 12x - 16y = 21}}} Now group the x-terms together and the y-terms together.
{{{(x^2 - 12x) + (y^2 - 16y) = 21}}} Next, complete the square in x by adding the square of half the x-coefficient to both sides of the equation. {{{(-12/2)^2 = 36}}}, and the same for the y {{{(-16/2)^2 = 64}}} 
{{{(x^2 - 12x + 36) + (y^2 - 16y + 64) = 21 + 36 + 64}}} Simplify.
{{{(x^2 - 12x + 36) + (y^2 - 16y + 64) = 121}}} Factor the x-group and the y-group.
{{{(x - 6)^2 + (y - 8)^2 = 121}}}  Take the square root of 121 = 11
{{{(x - 6)^2 + (y - 8)^2 = (11)^2}}} Compare with the standard form for a circle with centre at (h, k) and radius, r.
{{{(x - h)^2 + (y - k)^2 = r^2}}}

The circle centre is at (6, 8) and the radius is 11