$Find the Intersection Points on the Unit Circle$

Consider the unit circle defined by the equation $x^2 + y^2 = 1$ and the line with the equation $y = mx + c$.

To find the intersection points, substitute $y = mx + c$ into the unit circle’s equation:

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

Expand and simplify:

$x^2 + m^2x^2 + 2mxc + c^2 = 1$

Combine like terms:

$ (1 + m^2)x^2 + 2mxc + c^2 – 1 = 0 $

This is a quadratic equation in $x$:

$ Ax^2 + Bx + C = 0 $

where $A = 1 + m^2$, $B = 2mc$, and $C = c^2 – 1$.

Solve for $x$ using the quadratic formula:

$ x = \frac{-B \pm \sqrt{B^2 – 4AC}}{2A} $

Substitute the values of $A$, $B$, and $C$:

$ x = \frac{-2mc \pm \sqrt{(2mc)^2 – 4(1 + m^2)(c^2 – 1)}}{2(1 + m^2)} $

Simplify further to find $x$, then use $y = mx + c$ to find the $y$ values of the intersection points.

Given the unit circle $x^2 + y^2 = 1$ and the line $y = mx + c$, we need to find their intersection points.

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

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

Expand and combine terms:

$x^2 + m^2x^2 + 2mxc + c^2 = 1$

Group and simplify:

$ (1 + m^2)x^2 + 2mxc + c^2 – 1 = 0 $

Use the quadratic formula to solve for $x$:

$ x = frac{-2mc pm sqrt{4m^2c^2 – 4(1 + m^2)(c^2 – 1)}}{2(1 + m^2)} $

Simplify the expression under the square root:

$ x = frac{-2mc pm sqrt{4m^2c^2 – 4c^2 – 4m^2 + 4}}{2(1 + m^2)} $

Further simplify to find $x$, and then use $y = mx + c$ to find corresponding $y$ values.

Given $x^2 + y^2 = 1$ and $y = mx + c$:

Substitute $y$:

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

Combine terms:

$ (1 + m^2)x^2 + 2mxc + c^2 – 1 = 0 $

Use the quadratic formula:

$ x = frac{-2mc pm sqrt{4m^2c^2 – 4(1 + m^2)(c^2 – 1)}}{2(1 + m^2)} $

Simplify to find $x$ and use $y = mx + c$ for $y$.

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