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

To find the equation of the tangent line to the unit circle at $(1, 0)$, we first recognize that the unit circle has the equation:

$ x^2 + y^2 = 1 $

The slope of the tangent line at any point $(x_0, y_0)$ on the circle can be found using implicit differentiation:

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

At $(1, 0)$, this simplifies to:

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

Therefore,

$ \x0crac{dy}{dx} = 0 $

The slope is

txt1

txt1

txt1

$, so the equation of the tangent line is:

$ y = 0 $

The unit circle is given by the equation:

$ x^2 + y^2 = 1 $

Implicit differentiation gives:

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

At $(1, 0)$:

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

So,

$ x0crac{dy}{dx} = 0 $

The tangent line equation is:

$ y = 0 $

The unit circle is:

$ x^2 + y^2 = 1 $

Implicit differentiation yields:

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

At $(1, 0)$,

$ x0crac{dy}{dx} = 0 $

The tangent line is:

$ y = 0 $

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