#### Answer

$(4+\sqrt{7},2)$
$(4-\sqrt{7},2)$

#### Work Step by Step

A circle has the form:
$(x-h)^2+(y-k)^2=r^2$ with center $(h,k)$ and radius $r$
We have the center $(4,5)$ and radius $4$, so:
$(x-4)^{2}+(y-5)^{2}=4^{2}$
To find the intersection between the circle and the line $y=2$, we plug in $y=2$ into the circle equation and solve for $x$:
$(x-4)^{2}+(2-5)^{2}=4^{2}$
$(x-4)^{2}+(-3)^{2}=4^{2}$
$(x-4)^{2}=16-9$
$(x-4)^{2}=7$
$x-4=\pm\sqrt{7}$
$x=4\pm\sqrt{7}$
Since we know that $y=2$, the two intersection points must be:
$(4+\sqrt{7},2)$
$(4-\sqrt{7},2)$