Find the equation of the tangent line to the unit circle at $(1, 0)$

The unit circle is given by the equation:

$ x^2 + y^2 = 1 $

To find the equation of the tangent line at $(1, 0)$, we first find the slope of the tangent. Differentiate the equation implicitly with respect to $x$:

$ 2x + 2y \x0crac{dy}{dx} = 0 $

At the point $(1, 0)$, substitute $x = 1$ and $y = 0$:

$ 2(1) + 2(0) \x0crac{dy}{dx} = 0 $

So, the slope $\x0crac{dy}{dx}$ at $(1, 0)$ is

txt1

txt1

txt1

$. The equation of the tangent line using point-slope form is:

$ y – 0 = 0(x – 1) $

Therefore, the equation is:

$ y = 0 $

The unit circle is given by:

$ x^2 + y^2 = 1 $

Differentiating implicitly with respect to $x$:

$ 2x + 2y x0crac{dy}{dx} = 0 $

At $(1, 0)$,

$ 2(1) + 2(0) x0crac{dy}{dx} = 0 $

The slope $x0crac{dy}{dx}$ is

txt2

txt2

txt2

$, so the tangent line is:

$ y = 0 $

The unit circle is $ x^2 + y^2 = 1 $. Differentiating:

$ 2x + 2y x0crac{dy}{dx} = 0 $

At $(1, 0)$:

$ 2(1) + 2(0) x0crac{dy}{dx} = 0 $

The slope is

txt3

txt3

txt3

$, hence:

$ y = 0 $

Start Using PopAi Today