Question 202522
Well, in this type of problem (identifying curves from their equations), it is helpful to recognise the standard form of the equation for a circle with its centre at (h, k) and having a radius of r:
{{{(x-h)^2+(y-k)^2 = r^2}}} 
Do you see the similarity between this and the given equation?
{{{(x+2)^2+(y-2)^2 = 36}}}: h = -2, k = 2, and {{{r^2 = 36}}} so that {{{r = 6}}}
Can you take it from here?
By the way, you do not have to solve for x or y to answer the question.
But you will have to solve for y to graph the equation.
Here's how you would do this:
{{{(x+2)^2+(y-2)^2 = 36}}} Subtract {{{(x+2)^2}}} from both sides of the equation.
{{{(y-2)^2 = -(x+2)^2+36}}} Now take the square root of both sides.
{{{y-2 = sqrt(-(x+2)^2+36)}}} Add 2 to both sides to isolate the y.
{{{y = 2+-sqrt(-(x+2)^2+36)}}} You can simplify the radicand (the quantity under the radical sign) a bit.
{{{y = 2+-sqrt(-x^2-4x+32)}}}
Notice that you really have two answers here:
{{{y = 2+sqrt(-x^2-4x+32)}}} and {{{y = 2-sqrt(-x^2-4x+32)}}} and, to graph this, you would have to graph both parts, one for the upper half of the circle and the other for the lower half.
{{{graph(400,400,-10,5,-5,10,2+sqrt(-x^2-4x+32),2-sqrt(-x^2-4x+32))}}}