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
`(\mathrm dy)/(\mathrm dx)` with
`x=x_0` and
`y=y_0`. By the chain rule (and this is where we use implicit differentiation),

`(\mathrm dy)/(\mathrm dx)=((\mathrm dy)/(\mathrm dt))/((\mathrm dx)/(\mathrm dt))=(R\cos t)/(-R\sin t)=-(\cos t)/(\sin t)`

At the point
`P`, we have

`x_0=a+R\cos t\implies\cos t=\frac{x_0-a}R`

`y_0=b+R\sin t\implies\sin t=\frac{y_0-b}R`

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

`(\mathrm dy)/(\mathrm dx)=-\frac{\frac{x_0-a}R}{\frac{y_0-b}R}=-(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

`(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

`-\frac1{(b-y_0)/(a-x_0)}=-(a-x_0)/(b-y_0)=-(x_0-a)/(y_0-b)`

which is exactly the slope of the tangent line.