Question 64588
Find the center and radius of the circle whose equation is:
{{{x^2+y^2-6x-2y+5 = 0}}} Yes, you first subtract 5 from both sides.
{{{x^2+y^2-6x-2y = -5}}} Now group the x-terms together and the y-terms together.
{{{(x^2-6x) + (y^2-2y) = -5}}} Next, complete the squares in the x-terms and in the y-terms.
{{{(x^2-6x+9) + (y^2-2y+1) = -5+10}}} Now factor the left side.
{{{(x-3)^2 + (y-1)^2 = 5}}} Compare this with standard form for a circle with centre at (h, k) and radius r.
{{{(x-h)^2 + (y-k)^2 = r^2}}}
From this comparison, you can see that the centre of your circle is at (3, 1) and the radius{{{r = sqrt(5)}}}