Calculate the cartesian coordinates of the intersection points of the unit circle and the line $ y = 2x + 1 $

First, the equation of the unit circle is:

$ x^2 + y^2 = 1 $

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

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

Expand and simplify:

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

Combine like terms:

$ 5x^2 + 4x = 0 $

Factor out:

$ x(5x + 4) = 0 $

Then, $ x = 0 $ or $ x = -\x0crac{4}{5} $.

For $ x = 0 $:

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

So, one intersection point is $ (0, 1) $.

For $ x = -\x0crac{4}{5} $:

$ y = 2(-\x0crac{4}{5}) + 1 = -\x0crac{8}{5} + 1 = -\x0crac{3}{5} $

So, the other intersection point is $ (-\x0crac{4}{5}, -\x0crac{3}{5}) $.

The cartesian coordinates of the intersection points are $ (0, 1) $ and $ (-\x0crac{4}{5}, -\x0crac{3}{5}) $.

First, substitute $ y = 2x + 1 $ into the unit circle equation $ x^2 + y^2 = 1 $:

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

Expand:

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

Simplify:

$ 5x^2 + 4x = 0 $

Factor out:

$ x(5x + 4) = 0 $

Then, $ x = 0 $ or $ x = -x0crac{4}{5} $.

For $ x = 0 $:

$ y = 1 $

For $ x = -x0crac{4}{5} $:

$ y = -x0crac{3}{5} $

The intersection points are $ (0, 1) $ and $ (-x0crac{4}{5}, -x0crac{3}{5}) $.

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

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

Simplify:

$ 5x^2 + 4x = 0 $

Solve for $ x $:

$ x = 0 $ or $ x = -x0crac{4}{5} $

Then $ y = 1 $ or $ y = -x0crac{3}{5} $.

The intersection points are $ (0, 1) $ and $ (-x0crac{4}{5}, -x0crac{3}{5}) $.

Start Using PopAi Today