The standard equation of a circle with center (h,k) and radius r is given by:

\(\displaystyle{\left({x}−{h}\right)}^{{2}}+{\left({y}−{k}\right)}^{{2}}={r}^{{2}}\)

Using \((h,k)=(0,0)\) and \((x,y)=(0,2)\), we solve for r2:

\((0-0)2+(2-0)2=r2\)

\(4=r2\)

So, the equation of the circle is:

\(\displaystyle{x}^{{2}}+{y}^{{2}}={4}\)

The point (1,\(\sqrt{3}\)) lies on the circle if it satisfies the equation so we check:

\(\displaystyle{1}^{{2}}+{\left(√{3}\right)}^{{2}}=?{4}\)

\(1+3=?4\)

\(4=4\)

Since it satisfies the equation, then \(\displaystyle{\left({1},√{3}\right)}\) lies on the circle.