Find the coordinates of the point where the line $y = 2x$ intersects the unit circle.

First, let’s write the equation of the unit circle:

$x^2 + y^2 = 1.$

Since $y = 2x$, we can substitute $2x$ for $y$ in the unit circle equation:

$x^2 + (2x)^2 = 1.$

This simplifies to:

$x^2 + 4x^2 = 1$

$5x^2 = 1$

$x^2 = \frac{1}{5}$

$x = \pm \frac{1}{\sqrt{5}}$

Substituting these values back into $y = 2x$, we get:

$y = 2(\frac{1}{\sqrt{5}}) = \frac{2}{\sqrt{5}}$

$y = 2(-\frac{1}{\sqrt{5}}) = -\frac{2}{\sqrt{5}}$

Hence, the points of intersection are:

$ \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)$ and $ \left( -\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}} \right).$

The equation of the unit circle is given by:

$x^2 + y^2 = 1$

The line equation is:

$y = 2x$

Substitute $2x$ into the circle equation:

$x^2 + (2x)^2 = 1$

Simplify to:

$x^2 + 4x^2 = 1$

$5x^2 = 1$

$x^2 = frac{1}{5}$

$x = pm frac{1}{sqrt{5}}$

So, $y = frac{2}{sqrt{5}}$ or $y = -frac{2}{sqrt{5}}$

The intersection points are:

$ left( frac{1}{sqrt{5}}, frac{2}{sqrt{5}}
ight)$ and $ left( -frac{1}{sqrt{5}}, -frac{2}{sqrt{5}}
ight)$

For the unit circle $x^2 + y^2 = 1$, and the line $y = 2x$:

Substitute $y = 2x$ into $x^2 + y^2 = 1$:

$x^2 + 4x^2 = 1$

$5x^2 = 1$

$x = pm frac{1}{sqrt{5}}$

Thus, $y = frac{2}{sqrt{5}}$ or $y = -frac{2}{sqrt{5}}$

Intersection points: $ left( frac{1}{sqrt{5}}, frac{2}{sqrt{5}}
ight)$ and $ left( -frac{1}{sqrt{5}}, -frac{2}{sqrt{5}}
ight).$

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