## College Algebra (11th Edition)

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