Given a unit circle, find the equation of a tangent line at a point $(a,b)$ on the circle

To find the equation of the tangent line to the unit circle at the point $(a,b)$, recall that the unit circle is given by:

$ x^2 + y^2 = 1 $

Since the radius at the point $(a,b)$ is perpendicular to the tangent, the slope of the radius is:

$ m_r = \x0crac{b}{a} $

Thus, the slope of the tangent line, being the negative reciprocal, is:

$ m_t = -\x0crac{a}{b} $

Using the point-slope form of a line, the tangent line equation is:

$ y – b = -\x0crac{a}{b}(x – a) $

Simplifying, we get:

$ y = -\x0crac{a}{b}x + \x0crac{a^2 + b^2}{b} $

Since $(a,b)$ lies on the unit circle, we know:

$ a^2 + b^2 = 1 $

So the tangent line equation simplifies to:

$ y = -\x0crac{a}{b}x + \x0crac{1}{b} $

To determine the tangent line to the unit circle at $(a,b)$, where $a^2 + b^2 = 1$, we start with the unit circle equation:

$ x^2 + y^2 = 1 $

The slope of the radius at $(a,b)$ is:

$ x0crac{b}{a} $

Thus, the slope of the tangent line is:

$ -x0crac{a}{b} $

Using the point-slope form:

$ y – b = -x0crac{a}{b}(x – a) $

The tangent line equation simplifies to:

$ y = -x0crac{a}{b}x + x0crac{1}{b} $

The unit circle is $x^2 + y^2 = 1$. The tangent line at $(a,b)$ has slope $-a/b$. Hence, its equation is:

$ y – b = -x0crac{a}{b}(x – a) $

So, $ y = -x0crac{a}{b}x + x0crac{1}{b} $

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