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 $OP$.
Sunday, November 26, 2017
Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 43
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