Answer
Intersection points: $(\displaystyle \frac{1}{2},-\frac{\sqrt{3}}{2})$ and $(\displaystyle \frac{1}{2},\frac{\sqrt{3}}{2})$
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Work Step by Step
See attached image (desmos.com).
To find the intersection points algebraically,
Expand the square in the second equation.
$x^{2}-2x+1+y^{2}=1$
Now, substitute $x^{2}+y^{2}$ on the LHS with $1$
$1-2x+1=1$
$2-2x=1$
$1=2x$
$x=\displaystyle \frac{1}{2}$
Back-substitute:
$(\displaystyle \frac{1}{2})^{2}+y^{2}=1$
$y^{2}=1-\displaystyle \frac{1}{4}$
$y^{2}=\displaystyle \frac{3}{4}$
$y=\displaystyle \pm\frac{\sqrt{3}}{2}$
Intersection points: $(\displaystyle \frac{1}{2},-\frac{\sqrt{3}}{2})$ and $(\displaystyle \frac{1}{2},\frac{\sqrt{3}}{2})$
