Answer
$P(\frac{2\sqrt{5}}{5},\frac{\sqrt{5}}{5})$
Work Step by Step
We have the following equation for the unit circle:
$x^{2}+y^{2}=1$, so we plug in $P(x,\frac{\sqrt{5}}{5})$,
$x^{2}+(\frac{\sqrt{5}}{5})^{2}=1$
$ x^{2}=1 - \frac{5}{25}= \frac{20}{25}$
$x = \frac{2\sqrt{5}}{5}$ or $x = - \frac{2\sqrt{5}}{5}$ but the x-coordinate is positive so $x = \frac{2\sqrt{5}}{5}$
