Question 1108989
here's the graph.


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the dashed lines are only there to allow me to show you the coordinate points of interest.


you can see that:


the center is at (0,2).


the value of x goes from -5 to 5.


that's the domain.


the value of y goes from -3 to 7.


that's the range.


the radius of the circle is 5.


you can see that from the graph.


you can also determine all of this from the equation itself, without looking at the graph.


the standard form of the equation of a circle is (x-h)^2 + (y-k)^2 = r^2.


(h,k) is the center of the circle.


r is the radius of the circle.


your equation is x^2 + (y-2)^2 = r^2


that makes h = 0 and k = 2, which means the center of the circle is at (h,k) = (0,2).


the radius of the circle is sqrt(r^2) which gets you plus or minus 5.


the radius has to be positive, so the radius is 5.


from the center and the radius, you can find the domain and the range.


to find the domain, add and subtract the radius from h.


since h = 0, the domain is all values of x from -5 to 5.


to find the range, add and subtract the radius from k.


since k = 2, the range is all value of y from -3 to 7.


the graph confirms this to be true.


if you have the equation of the circle in standard form, then finding domain and range from the equation is fairly simple, as shown above.


if the equation is not in standard form, then you need to convert it to standard form by using the completing the square method.


here's a reference on how to do that.


<a href = "http://www.purplemath.com/modules/sqrcircle.htm" target = "_blank">http://www.purplemath.com/modules/sqrcircle.htm</a>