Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 43

By using implicit differentiation, show that any tangent line at a point $P$ to a circle with center $O$ is perpendicular to the radius $OP$.

Assuming that the circle is centered at origin, its equation is

$x^ 2 + y^2 = r^2$

Taking the derivative of the curve implicitly we have,


$\begin{equation} \begin{aligned} 2x + 2y \frac{dy}{dx} =& 0 \\ \\ \frac{dy}{dx} =& \frac{-x}{y} \end{aligned} \end{equation}$


Thus the slope of the tangent at $P(x_1, y_1)$ is $\displaystyle \frac{-x_1}{y_1}$

Also, the slope of the radius connecting the origin and point $P(x_1, y_1)$ can be completed as $\displaystyle m = \frac{y_1 - 0}{x_1 - 0}$

But, we know that the slope of the normal line is equal to negative reciprocal of the slope of the tangent line so..


$\begin{equation} \begin{aligned} m_T =& \frac{-1}{m_N} \\ \\ \frac{-x_1}{y_1} =& \frac{-1}{m_N} \\ \\ m_N =& \frac{y_1}{x_1} \qquad \qquad \text{which equals the slope of the radius} \end{aligned} \end{equation}$



Therefore, it shows that any tangent line at a point $P$ to a circle with center $O$ is perpendicular to the radius $OP$.