June 2019

The tangent line of a circle can be defined as a line that intersects the circle at one point only.

Put a circle in the rectangular coordinate system.

Let $(x_0, y_0)$ be a point on a circle. The tangent line at $(x_0, y_0)$ is a line intersects the circle at $(x_0, y_0)$ only.

Let’s first find a function $y=kx+m$ that represents the line.

From circle’s equation $x^2+y^2=r^2$ , we have

$y^2=r^2-x^2$

Since the line intersects the circle at $(x_0, y_0)$ only,

$r^2-x^2=(kx+m)^2$

has only one solution.

That means

$k^2x^2+x^2+2kmx+m^2-r^2 =0$

has only one solution. i.e., its discriminant

$(2km)^2-4(k^2+1)(m^2-r^2)=0\quad\quad\quad(1)$

By the definition of slope,

$kx+m-y_0 = k(x-x_0)$ .

It follows that

$m =y_0-kx_0\quad\quad\quad(2)$

Substitute (2) into (1) and solve for $k$ gives

$k = \frac{-x_0}{y_0}\quad\quad\quad(3)$

The slope of line connecting $(0, 0)$ and $(x_0, y_0)$ where $x_0 \neq 0$ is $\frac{y_0}{x_0}$ .

Since $\frac{-x_0}{y_0}\cdot \frac{y_0}{x_0} = -1$ , the tangent line is perpendicular to the line connecting $(0, 0)$ and $(x_0, y_0)$ .

Substitute (3) into $y = k x +m$ , we have

$y=-\frac{x_0}{y_0} x + m\quad\quad\quad(4)$ .

The fact that the line intersects the circle at $(x_0, y_0)$ means

$y_0 = -\frac{x_0^2}{y_0} + m$

$y_0^2=-x_0^2+ my_0$ .

Hence,

$m =\frac{x_0^2+y_0^2}{y_0} = \frac{r^2}{y_0}$ .

It follows that by (4),

$x_0 x +y_0 y = r^2\quad\quad\quad(5)$

(5) is derived under the assumption that $y_0 \neq 0$ . However, by letting $y_0 =0$ in (5), we obtain two tangent lines that can not be expressed in the form of $y=kx+m$ :

$x=-r, x=r$

The tangent line of a quadratic function at $(x_0, y_0)$ is a line $y=kx+m$ that intersects $y=ax^2+bx+c$ at $(x_0, y_0=ax_0^2+bx_0+c)$ only.

The presence of function $y=kx+m$ immediately excludes the vertical line $x=x_0$ which also intersects $y=ax^2+bx+c$ at $(x_0, ax_0^2+bx_0+c)$ only (see Fig. 1).