Answer
Intersection points: $(\displaystyle \frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5})$ and $(-\displaystyle \frac{\sqrt{5}}{5},-\frac{2\sqrt{5}}{5})$
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Work Step by Step
See attached image (desmos.com).
To find the intersection points algebraically,
substitute $y=2x $ into the second equation:
$x^{2}+(2x)^{2}=1$
$x^{2}+4x^{2}=1$
$5x^{2}=1$
$x^{2}=\displaystyle \frac{1}{5}$
$x=\displaystyle \pm\sqrt{\frac{1}{5}}=\pm\frac{\sqrt{5}}{5}$
Back-substitute:
$x=\displaystyle \frac{\sqrt{5}}{5}\quad \Rightarrow\quad y= \frac{2\sqrt{5}}{5}$
$x=-\displaystyle \frac{\sqrt{5}}{5}\quad \Rightarrow\quad y= -\frac{2\sqrt{5}}{5}$
Intersection points: $(\displaystyle \frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5})$ and $(-\displaystyle \frac{\sqrt{5}}{5},-\frac{2\sqrt{5}}{5})$
