Find the intersection points of the unit circle and the line $y = x - 1$.

To find the intersection points, we need to solve the system of equations formed by the unit circle equation and the given line equation.

The unit circle equation is:

$x^2 + y^2 = 1$

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

$x^2 + (x – 1)^2 = 1$

Expand and simplify:

$x^2 + x^2 – 2x + 1 = 1$

Combine like terms:

$2x^2 – 2x + 1 = 1$

Simplify further:

$2x^2 – 2x = 0$

Factor out the common term:

$2x(x – 1) = 0$

Set each factor to zero:

$x = 0$

$x = 1$

For $x = 0$:

$y = 0 – 1 = -1$

For $x = 1$:

$y = 1 – 1 = 0$

Thus, the intersection points are $(0, -1)$ and $(1, 0)$.

Given the unit circle equation:

$x^2 + y^2 = 1$

And the line equation:

$y = x – 1$

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

$x^2 + (x – 1)^2 = 1$

Simplify and solve the quadratic equation:

$x^2 + x^2 – 2x + 1 = 1$

$2x^2 – 2x = 0$

$2x(x – 1) = 0$

Thus, $x = 0$ or $x = 1$.

For $x = 0$:

$y = 0 – 1 = -1$

For $x = 1$:

$y = 1 – 1 = 0$

The intersection points are $(0, -1)$ and $(1, 0)$.

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

$x^2 + (x – 1)^2 = 1$

$2x^2 – 2x = 0$

$2x(x – 1) = 0$

$x = 0$, $y = -1$

$x = 1$, $y = 0$

Intersection points: $(0, -1)$ and $(1, 0)$.

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