Question

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

Answer 1

A circle centered at $O(a,b)$ with radius $R$ (the length of $OP$) has equation

$(x-a)^2+(y-b)^2=R^2$

which can be parameterized by

$\vec c(t)=\langle x(t),y(t)\rangle=\langle a+R\cos t,b+R\sin t\rangle$

with $0\le t\le2\pi$.

The tangent line to $\vec c(t)$ at a point $P(x_0,y_0)$ is $\frac{\mathrm dy}{\mathrm dx}$ with $x=x_0$ and $y=y_0$. By the chain rule (and this is where we use implicit differentiation),

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{R\cos t}{-R\sin t}=-\dfrac{\cos t}{\sin t}$

At the point $P$, we have

$x_0=a+R\cos t\implies\cos t=\dfrac{x_0-a}R$

$y_0=b+R\sin t\implies\sin t=\dfrac{y_0-b}R$

so that the slope of the line tangent to the circle at $P$ is

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{\frac{x_0-a}R}{\frac{y_0-b}R}=-\dfrac{x_0-a}{y_0-b}$

Meanwhile, the slope of the line through the center $O(a,b)$ and the point $P(x_0,y_0)$ is

$\dfrac{b-y_0}{a-x_0}$

Recall that perpendicular lines have slopes that are negative reciprocals of one another; taking the negative reciprocal of this slope gives

$-\dfrac1{\frac{b-y_0}{a-x_0}}=-\dfrac{a-x_0}{b-y_0}=-\dfrac{x_0-a}{y_0-b}$

which is exactly the slope of the tangent line.