Question 204734
Start with te general form of the equation of a circle with center at (h, k) and radius r.
{{{(x-h)^2+(y-k)^2 = r^2}}}
The given equation is:
{{{x^2-10x+y^2-8y = 8}}} First, group the terms as shown:
{{{(x^2-10x)+(y^2-8y) = 8}}} Now you need to "complete the squares" in both the x-terms and the y-terms. You do this by adding the square of half the x-coefficient and the square of half the y=coefficient to both sides of the equation.
The square of half the x-coefficient is: {{{(-10/2)^2 = 25}}} and the square of half the y-coefficient is: {{{(-8/2)^2 = 16}}} so...
{{{(x^2-10+25)+(y^2-8y+16) = 8+25+16}}} On the left side, factor the x-group and factor the y-group.
{{{(x-5)^2+(y-4)^2 = 29}}} Now compare this with the general form from above:
{{{(x-h)^2+(y-k)^2 = r^2}}}
You can see that:
{{{h = 5}}}, {{{k = 4)}}}, and {{{r^2 = 49}}}
So the center (h, k) is at (5, 4) and the radius, {{{r = sqrt(49)}}} or {{{r = 7}}}