Hi all ---

I'm posting this question on geometry here because I'm not sure if you need calculus. But even if you don't, I'd like to know if I can use calculus to do the question.

Question with solution:

Now, I understand perfectly how to complete the square. So when I did this problem, I graphed it and saw that the point on the circle closest to $\displaystyle (0, 0) $ had to be in the 1st quadrant. This is because I can see that the origin is in the upper-right "quarter" of the circle, if I separate the circle into equal 4s.

But I'm not sure how to find the specific point in the 1st quadrant? The solution somehow knows that $\displaystyle (3,4) $ is the answer, but it doesn't explain how it got it?

Also, I know from calculus that ---

$\displaystyle \text{distance} = D = \sqrt{(x - a)^2 + (y - b)^2} $ where $\displaystyle x, y $ must be on the circle.

But I can maximize the square instead ---

$\displaystyle d = D^2 = (x - a)^2 + (y - b)^2} $ where $\displaystyle x, y $ must be on the circle.

But this doesn't seem to help, because it looks too complicated to sub in the equation of the circle into $\displaystyle d$?

Thanks ---