Monthly Archives: June 2019

Constructing the tangent line of circle without calculus

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

or

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

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Constructing the tangent line of quadratic without calculus

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).

Fig. 1

Let’s find k.

Line y = kx+m intersects y=ax^2+bx+c at (x_0, ax_0^2+bx_0+c) only means quadratic equation

ax^2+bx +c =kx +m

has only one solution. That is, the discriminant of ax^2+bx+c-kx-m =0 is zero:

(b-k)^2-4a(c-m) = 0\quad\quad\quad(1)

Fig. 2

And, by the definition of slope (see Fig. 2),

(x-x_0)k = (kx+m)-(ax_0^2+bx_0+c).

It follows that

m = (ax_0^2+b_0+c)-x_0 k\quad\quad\quad(2)

Substituting (2) into (1), we have

(b-k)^2-4a(c-((a_0 x^2+b x_0 + c)-x_0 k)=0.

Solve it for k gives

k = 2 a x_0 +b.

Fig. 3