Find the points of intersection between the unit circle and the line $y = 2x + 1$

To find the points of intersection, we can substitute $y = 2x + 1$ into the equation of the unit circle, which is $x^2 + y^2 = 1$.

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

Expanding the equation:

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

Combining like terms:

$5x^2 + 4x + 1 = 1$

Simplifying:

$5x^2 + 4x = 0$

Factoring the equation:

$x(5x + 4) = 0$

So $x = 0$ or $x = -\frac{4}{5}$.

When $x = 0$, $y = 1$.

When $x = -\frac{4}{5}$, $y = 2(-\frac{4}{5}) + 1 = -\frac{8}{5} + 1 = -\frac{3}{5}$.

Thus, the points of intersection are $(0, 1)$ and $(-\frac{4}{5}, -\frac{3}{5})$.

Intersecting the line $y = 2x + 1$ with the unit circle $x^2 + y^2 = 1$, substitute $y$ in the circle’s equation:

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

Expanding it:

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

Combining terms:

$5x^2 + 4x + 1 = 1$

Simplifying further:

$5x^2 + 4x = 0$

Factoring out $x$:

$x(5x + 4) = 0$

Solving for $x$ gives $x = 0$ or $x = -frac{4}{5}$.

Calculating $y$ for $x = 0$, $y = 1$.

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

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

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

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

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

Expanding:

$5x^2 + 4x + 1 = 1$

Simplifying:

$5x^2 + 4x = 0$

Factoring:

$x(5x + 4) = 0$

So $x = 0$ or $x = -frac{4}{5}$.

For $x = 0$, $y = 1$. For $x = -frac{4}{5}$, $y = -frac{3}{5}$.

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

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