My standard-form equation is (x+3)^2+(y+2)^2=4. Do I find the radius by doing the square root of four? I just wanted to check on this. Based on the example in my book, it seemed like this is the way to do this, but I wanted to make sure, because they didn't come right out and say that. Thanks for your help!!
yes, in standard form, the radius is the square root of the right hand side.
To find the radius of a circle given its standard-form equation, you'll need to compare it to the equation of a circle in standard form, which is of the form (x-h)^2 + (y-k)^2 = r^2.
In your equation (x+3)^2 + (y+2)^2 = 4, notice that the right side of the equation is not r^2. It is actually equal to 4, not r^2. Therefore, simply taking the square root of 4 would not give you the radius.
To find the radius, you need to solve for r^2. In this case, since 4 is already on the right side of the equation, it means that r^2 = 4. So the radius squared is 4, and taking the square root of 4 gives you the radius, which is 2.
Therefore, the radius of the circle described by the equation (x+3)^2 + (y+2)^2 = 4 is 2.
In general, when given a standard-form circle equation, if the right side is a perfect square, you can take the square root to find the radius. But if it's not a perfect square, you need to first isolate r^2 and then take the square root to determine the radius.