Find the points where the line $ y = mx + c $ intersects the unit circle

To find the points of intersection between the line $ y = mx + c $ and the unit circle $ x^2 + y^2 = 1 $, we substitute the expression for y into the circle’s equation:

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

Expanding and simplifying:

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

Combining like terms:

$ (1 + m^2)x^2 + 2mcx + c^2 – 1 = 0 $

This is a quadratic equation in x. To solve for x, we use the quadratic formula:

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

Simplifying under the square root and the denominator:

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

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

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

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

Thus, we find the x-coordinates of the intersection points as:

$ x_1 = \frac{-mc + \sqrt{m^2 + 1 – c^2}}{1 + m^2}, x_2 = \frac{-mc – \sqrt{m^2 + 1 – c^2}}{1 + m^2} $

The corresponding y-coordinates are found by substituting these x-values back into the line equation $ y = mx + c $.

To determine the intersection points of the line $ y = mx + c $ with the unit circle $ x^2 + y^2 = 1 $, substitute $ y = mx + c $ into the circle’s equation:

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

This becomes:

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

Combining terms:

$ (1 + m^2)x^2 + 2mcx + c^2 – 1 = 0 $

Using the quadratic formula:

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

Simplifying:

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

We get:

$ x_1 = frac{-mc + sqrt{m^2 + 1 – c^2}}{1 + m^2}, x_2 = frac{-mc – sqrt{m^2 + 1 – c^2}}{1 + m^2} $

Then substituting $ x_1 $ and $ x_2 $ into $ y = mx + c $ to find the corresponding y-values.

The points of intersection can be found by solving the system of equations given by the line $ y = mx + c $ and the unit circle $ x^2 + y^2 = 1 $. Substituting $ y $ from the line equation into the circle’s equation gives:

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

Expanding and simplifying, we get a quadratic equation in terms of $ x $:

$$ (1 + m^2)x^2 + 2mcx + (c^2 – 1) = 0 $$

Solving this equation for $ x $ will give the x-coordinates of the intersection points. The corresponding y-coordinates can be found by plugging these x-values into the line’s equation $ y = mx + c $.

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