Question 284792
This kind of equations, i.e. (x - a)^2 + (y - b)^2 = R^2 is a equation of a circle and its center is at (a,b) and the radius is R.
In your equation, 2x^2 + 2y^2 = 18 <-> x^2 + y^2 = 9 <-> (x-0)^2 + (y-0)^2 = 3^2
Therefore, it is a circle with its center at (0,0) and the radius length is 3.
So you know why it is C, right?

By the way, if the coefficients of x^2 and y^2 are equals (i.e. you always have the equations look like this: k(x-a)^2 + k(x-b)^2 = k*R^2 (k<>0), you can be sure that its graph is a circle, but when the coefficients are not equal, you have a  graph of an ellipse. 
Good luck.