Calculate the points of intersection of the unit circle and the line passing through the origin with a given slope $ m $

To find the points of intersection of the unit circle $x^2 + y^2 = 1$ and the line passing through the origin with slope $m$, we start with the equation of the line, $y = mx$.

Substituting $y = mx$ into the unit circle equation:

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

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

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

$x^2 = \frac{1}{1 + m^2}$

$x = \pm \frac{1}{\sqrt{1 + m^2}}$

Then, using $y = mx$,

$y = \pm \frac{m}{\sqrt{1 + m^2}}$

The points of intersection are:

$(\frac{1}{\sqrt{1 + m^2}}, \frac{m}{\sqrt{1 + m^2}})$ and $(\frac{-1}{\sqrt{1 + m^2}}, \frac{-m}{\sqrt{1 + m^2}})$

We need to determine the points of intersection of the unit circle $x^2 + y^2 = 1$ and the line $y = mx$. By substituting $y = mx$ in the unit circle equation:

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

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

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

$x^2 = frac{1}{1 + m^2}$

Taking the square root:

$x = pm frac{1}{sqrt{1 + m^2}}$

Using $y = mx$ we get:

$y = m cdot pm frac{1}{sqrt{1 + m^2}}$

The intersection points are:

$(frac{1}{sqrt{1 + m^2}}, frac{m}{sqrt{1 + m^2}})$ and $(frac{-1}{sqrt{1 + m^2}}, frac{-m}{sqrt{1 + m^2}})$

Given the unit circle $x^2 + y^2 = 1$ and the line $y = mx$, we substitute to find:

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

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

$x = pm frac{1}{sqrt{1 + m^2}}$

$y = pm frac{m}{sqrt{1 + m^2}}$

The intersection points are:

$(frac{1}{sqrt{1 + m^2}}, frac{m}{sqrt{1 + m^2}})$ and $(frac{-1}{sqrt{1 + m^2}}, frac{-m}{sqrt{1 + m^2}})$

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