Question 201163
Find the radius of the circle:
{{{x^2+y^2-12x+8y+43 = 0}}}
To do this, you need to change the given equation into the general form for a circle with center at (h, k) and radius, r,: {{{(x-h)^2+(y-k)^2 = r^2}}}.
You can do this by using the method of "completing the square". 
First, group the terms of the given equation as shown:
{{{(x^2-12x)+(y^2+8y)+43 = 0}}} Next, subtract 43 from both sides of the equation.
{{{(x^2-12x)+(y^2+8y) = -43}}} Now complete the square in the x-group and in the y-group by adding the square of half the x-coefficient and the square of half the y-coefficient to both sides of the equation.
{{{(x^2-12x+highlight(36))+(y^2+8y+highlight(16)) = -43+highlight(36)+highlight(16)}}} Now factor the x-trinomial and the y-trinomial and simplify the right side of the equation.
{{{(x-6)^2+(y+4)^2 = 9}}} Compare this result with the standard form given above:
{{{(x-h)^2+(y-k)^2 = r^2}}}
You can see that the radius squared is {{{r^2 = 9}}} so the radius must be {{{highlight(r = 3)}}} and, incidentally, you now have the center of the circle, (h, k) and it's ({{{6}}},{{{-4}}})