Answer
$$P(\frac{2}{5}, \frac{\sqrt 21}{5})$$
Work Step by Step
Since $(\frac{2}{5}, [])$ is an $(x, y)$ coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$.
$$(\frac{2}{5})^{2}+(y)^{2}=1$$ $$\frac{2\times2}{5\times5}+y^{2}=1$$ $$\frac{4}{25}+y^{2}=1$$ $$y^{2}=1-\frac{4}{25}$$ $$y^{2}=\frac{25}{25}-\frac{4}{25}$$ $$y^{2}=\frac{21}{25}$$ $$\sqrt (y^{2})=\sqrt (\frac{21}{25})$$ $$y=±\frac{\sqrt 21}{5}$$ Since the coordinate needs to be in the 1st quadrant, the y value needs to be positive. Therefore, $y=\frac{\sqrt 21}{5}$ and $P(\frac{2}{5}, \frac{\sqrt 21}{5})$
