Question 187927
Find the center and the radius of the circle given by the equation:
{{{x^2-6x+y^2+10y = -30}}} First group the x- and y-variables as shown:
{{{(x^2-6x)+(y^2+10y) = -30}}} Now complete the squares in both the x- and y-groups by adding the square of half the x-coefficient and the square of half the y-coefficient to both sides of the equation.
{{{(x^2-6x+(-6/2)^2)+(y^2+10y+(10/2)^2) = -30+(-6/2)^2+(10/2)^2}}} Simplify this.
{{{(x^2-6x+9)+(y^2+10y+25) = -30+9+25}}} Factor the x-trinomial and factor the y-trinomial and simplify the right side.
{{{(x-3)^2 + (y+5)^2 = 4}}} Now compare this with the standard form of the equation of a circle with its center at (h, k) and whose radius is r.
{{{(x-h)^2 + (y-k)^2 = r^2}}} and you can readily see that:
{{{h = 3}}}
{{{k = -5}}}
{{{r = 2}}}
So the center of the circle is at (3, -5) and the radius is 2.