Answer
$\frac{(x-1)^2}{4}+\frac{(y-2)^2}{4}=1$
See graph.
Work Step by Step
1. Given center $(1,2)$, vertex $(1,4)$ and contain point $(1+\sqrt 3,3)$, we can identify a vertical major axis, $a=4-2=2$, $b^2+c^2=a^2=4$
2. Write the equation as $\frac{(x-1)^2}{b^2}+\frac{(y-2)^2}{a^2}=1$, use the point on curve to get $\frac{(1+\sqrt 3-1)^2}{b^2}+\frac{(3-2)^2}{4}=1$, thus $b=2$ and $c^2=4-4=0$, a circle.
3. Thus we have $\frac{(x-1)^2}{4}+\frac{(y-2)^2}{4}=1$
4. See graph.
