Question 534333
Compare the given equation
{{{(x-3)^2+(y+2)^2 = 9}}} with
{{{(x-h)^2+(y-k)^2 = r^2}}}} This is the general form of the equation of a circle with its center at (h,k) and a radius of r.
So you can easily identify the center and radius of the given equation as:
Center at (3, -2) and radius of 3.
To graph the equation, you must first solve the equation for y:
{{{(x-3)^2+(y+2)^2 = 9}}} Subtract {{{(x-3)^2}}} from both sides.
{{{(y+2)^2 = 9-(x-3)^2}}} Take the square root of both sides.
{{{y+2 = sqrt(9-(x-3)^2)}}} Subtract 2 from both sides.
{{{y = sqrt(9-(x-3)^2)-2}}} and {{{y = -sqrt(9-(x-3)^2)-2}}} Now graph these two equations.
{{{graph(400,400,-6,15,-15,6,sqrt(9-(x-3)^2)-2,-sqrt(9-(x-3)^2)-2)}}}