Determine the coordinates of the points where the unit circle intersects the line $ y = 2x + 1 $

First, recall the equation of the unit circle:

$ x^2 + y^2 = 1 $

Substitute $ y = 2x + 1 $ into the unit circle equation:

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

Expand and simplify the equation:

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

$ 5x^2 + 4x = 0 $

Factor the quadratic equation:

$ x(5x + 4) = 0 $

Thus, $ x = 0 $ or $ x = -\frac{4}{5} $. For $ x = 0 $:

$ y = 2(0) + 1 = 1 $

For $ x = -\frac{4}{5} $:

$ y = 2\left(-\frac{4}{5}\right) + 1 = -\frac{8}{5} + 1 = -\frac{3}{5} $

The intersection points are:

$ (0, 1) \: \text{and} \: \left(-\frac{4}{5}, -\frac{3}{5}\right) $

Start with the unit circle equation:

$ x^2 + y^2 = 1 $

Replace $ y $ with $ 2x + 1 $:

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

Expand:

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

Simplify:

$ 5x^2 + 4x = 0 $

Factor:

$ x(5x + 4) = 0 $

So, $ x = 0 : ext{or} : x = -frac{4}{5} $. Calculate $ y $ accordingly:

$ y = 1 : ext{for} : x = 0 $

$ y = -frac{3}{5} : ext{for} : x = -frac{4}{5} $

Intersection points:

$ (0, 1) : ext{and} : left(-frac{4}{5}, -frac{3}{5}
ight) $

Use the unit circle equation:

$ x^2 + y^2 = 1 $

Substitute $ y = 2x + 1 $:

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

Simplify:

$ 5x^2 + 4x = 0 $

Solve for $ x $:

$ x = 0 : ext{or} : x = -frac{4}{5} $

Find $ y $:

$ y = 1 : ext{and} : y = -frac{3}{5} $

Points:

$ (0, 1) : ext{and} : left(-frac{4}{5}, -frac{3}{5}
ight) $

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