The candidate should come up with something similar to the Midpoint circle algorithm also commonly referred to as Bresenham's Circle algorithm because of the similarities to his line algorithm.
Some other good questions regard reading code and detmining whether it not it will compile. If it will compile what will be the output, if it won't why not.
These questions shouldn't be a quiz of Library functions, just simple syntax. No complicated long lines of multiple functions crammed into a single line or anything mis-leading

The answer here is commonly referred to as Bresenham's Circle algorithm due to the similarities with his Line algorithm, but is actually called the Midpoint Circle algorithm.
I previously gave this answer in email to Shirley and wanted to elaborate further.
This can be difficult to come up with in an interview without memorizing, so to work toward it (which is what the interviewer wants to see) try this.
Really the interviewer is looking for you to move from x^2 + y^2 = R^2 to x^2 + y^2 - R^2 = 0.
You should calculate R^2 only once at function init. The toughest part is to get the y from without simply solving for y^2 then taking the square root. (sadly I drew blank on this in the interview and needless to say, I did not impress that interviewer).
You also don't want to calculate every point in the circle or draw the same point twice, so start by calling DrawPoint(center.x + r, center.y), DrawPoint(center.x - r, center.y), DrawPoint(center.x, center.y + r), and DrawPoint(center.x, center.y -r).
Then you only have to calculate the points for 1/8th of the circle.
First translate to 0,0.
Then iterate from the x value 1-r //this is 0-r+1 since we already calculated the first point
and iterate up to -r/2 //this is the 1/8th point halfway between 0 and 0-r
so that looks like this
for (int x = 1-r; x <= -r/2; x++)
//call the function that you wrote to solve for y from x^2 and your previously calculated rSquared value
//now remember that we are working from 0,0 so we have to shift our answers back to center and draw the Points that are easily mirrored from your answers
DrawPoint(new Point(x, y) + center)
DrawPoint(new Point(-x,y) + center)
DrawPoint(new Point(x, -y) + center)
DrawPoint(new Point(-x, -y) + center)
DrawPoint(new Point(x, y) + center)
DrawPoint(new Point(center - new Point(x,0))
DrawPoint(new Point(center + new Point(x,0))
DrawPoint(new Point(center - new Point(0, y))
DrawPoint(new Point(center + new Point(0, y))

or something like this, wife is rushing me, g2g.
I would also do this without all the new Point stuff to prevent local storage creation by just using the x and y values and calculating inside the parens or another efficient method for doing the same.

Have fun with the solving for y, it was really easy when I looked at it again, so i won't spoil it here, but once you look at Bresenham's you'll see what I mean

I knew I should've spent a little more time looking at that. sorry, the 8 points you want to draw are:
(x,y) + center
(y,x) + center
(-x,y) + center
(-y,x) + center
(x,-y) + center
(y, -x) + center
(-x,-y) + center and
(-y,-x) + center

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