Answer
$\displaystyle \{\ (\frac{\sqrt{31}}{4},\frac{7}{4}),\ (-\frac{\sqrt{31}}{4},\frac{7}{4})\ \}$
Work Step by Step
1. Eliminating:
Eliminate the $(y^{2})$ terms:
$\left\{\begin{array}{lll}
x^{2}+y^{2}=5 & , & /\times(-1),\\
x^{2}+(y-8)^{2}=41 & , & / add
\end{array}\right.$
$(y-8)^{2}-y^{2}=41-5$
2. Solving:
expand the square,
$y^{2}-16x+64-y^{2}=36$
$-16y+64=36\qquad/-64$
$-16y=-28\qquad/\div(-16)$
$y=\displaystyle \frac{-28}{-16}=\frac{7}{4}$
3. Back-substituting into $x^{2}+y^{2}=5$:
$x^{2}+\displaystyle \frac{49}{16}=5$
$x^{2}=\displaystyle \frac{80-49}{16}$
$x=\displaystyle \frac{\pm\sqrt{31}}{4}$
The solution set is
$\displaystyle \{\ (\frac{\sqrt{31}}{4},\frac{7}{4}),\ (-\frac{\sqrt{31}}{4},\frac{7}{4})\ \}$
