# Oh! Matryoshka!

Given polynomial $f(x) = a_0 + a_1 x+a_2 x^2 + ... + a_{n-1}x^{n-1}+a_n x^n$, we wish to evaluate integral

$\int \frac{f(x)}{(x-a)^p}\;dx, \quad p \in N^+\quad\quad\quad(1)$

When $p = 1$,

$\int \frac{f(x)}{x-a} \;dx= \int \frac{f(x)-f(a)+f(a)}{x-a}\;dx$

$= \int \frac{f(x)-f(a)}{x-a}\;dx + \int \frac{f(a)}{x-a}\;dx$

$=\int \frac{f(x)-f(a)}{x-a}\;dx + f(a)\cdot \log(x-a)$.

Since

$f(x) = a_0 + a_1x + a_2x^2 + ... + a_{n-1}x^{n-1} + a_n x^n$

and

$f(a) = a_0 + a_1 a + a_2 a^2 + ... + a_{n-1}a^{n-1} + a_n a^n$

It follows that

$f(x)-f(a) = a_1(x-a) + a_2(x^2-a^2) + ... + a_{n-1}(x^{n-1}-a^{n-1}) + a_n (x^n-a^n)$.

That is

$f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x^k-a^k)$

By the fact (see “Every dog has its day“) that

$x^k-a^k =(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}$,

we have

$f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}=(x-a)\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})$

or,

$\frac{f(x)-f(a)}{x-a}= \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\quad\quad\quad(2)$

Hence,

$\int\frac{f(x)}{x-a}\;dx = \int \sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\;dx + f(a)\log(x-a)$

$=\sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}\int x^{k-i}a^{i-1}\; dx)+ f(a)\log(x-a)$

i.e.,

$\int \frac{f(x)}{x-a} = \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a)$

Let us now consider the case when $p>1$:

$\int \frac{f(x)}{(x-a)^p}\; dx$

$=\int \frac{f(x)-f(a)+f(a)}{(x-a)^p}\;dx$

$=\int \frac{f(x)-f(a)}{(x-a)^p} + \frac{f(a)}{(x-a)^p}\;dx$

$=\int \frac{f(x)-f(a)}{(x-a)}\cdot\frac{1}{(x-a)^{p-1}} + \frac{f(a)}{(x-a)^p}\;dx$

$= \int \frac{f(x)-f(a)}{x-a}\cdot\frac{1}{(x-a)^{p-1}}\;dx + \int\frac{f(a)}{(x-a)^p}\; dx$

$\overset{(2)}{=}\int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}$

where

$g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})$, a polynomial of order $n-1$.

What emerges from the two cases of $p$ is a recursive algorithm for evaluating (1):

Given polynomial $f(x) = \sum\limits_{k=0}^{n} a_k x^k$,

$\int \frac{f(x)}{(x-a)^p} \;dx, \; p \in N^+= \begin{cases}p=1: \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a) \\p>1: \int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}, \\ \quad\quad\quad g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}). \end{cases}$

Exercise-1 Optimize the above recursive algorithm (hint: examine how it handles the case when $f(x)=0$)

# Integration of Trigonometric Expressions

We will introduce an algorithm for obtaining indefinite integrals such as

$\int \frac{(1+\sin(x))}{\sin(x)(1+\cos(x))}\;dx$

or, in general, integral of the form

$\int R(\sin(x), \cos(x))\;dx\quad\quad\quad(1)$

where $R$ is any rational function $R(p, q)$, with $p=\sin(x), q=\cos(x)$.

Let

$t = \tan(\frac{x}{2})\quad\quad(2)$

Solving (2) for $x$, we have

$x = 2\cdot\arctan(t)\quad\quad\quad(3)$

which provides

$\frac{dx}{dt} = \frac{2}{1+t^2}\quad\quad\quad(4)$

and,

$\sin(x) =2\sin(\frac{x}{2})\cos(\frac{x}{2})\overset{\cos^(\frac{x}{2})+\sin^2(\frac{x}{2})=1}{=}\frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{2\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

yields

$\sin(x) = \frac{2 t}{1+t^2}\quad\quad\quad(5)$

Similarly,

$\cos(x) = \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

gives

$\cos(x)=\frac{1-t^2}{1+t^2}\quad\quad\quad(6)$

We also have (see “Finding Indefinite Integrals” )

$\int f(x)\;dx \overset{x=\phi(t)}{=} \int f(\phi(t))\cdot\frac{d\phi(t)}{dt}\;dt$.

Hence

$\int R(\cos(x), \sin(x))\;dx \overset{(2), (4), (5), (6)}{=} \int R(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2})\cdot\frac{2}{1+t^2}\;dt$,

and (1) is reduced to an integral of rational functions in $t$.

Example-1 Evaluate $\int \csc(x)\;dx$.

Solution: $\csc(x) = \frac{1}{\sin(x)}\implies \int \csc(x)\;dx = \int \frac{1}{\sin(x)}\;dx$

$= \int \frac{1}{\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\;dt=\int\frac{1}{t}\;dt = \log(t) = \log(\tan(\frac{x}{2}))$.

Example-2 Evaluate $\int \sec(x)\;dx$.

Solution: $\sec(x) = \frac{1}{\cos(x)}\implies \int \sec(x)\; dx =\int \frac{1}{\cos(x)}\;dx$

$= \int \frac{1}{\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\; dt=\int \frac{2}{1-t^2}\;dt=\int \frac{2}{(1+t)(1-t)}\;dt=\int \frac{1}{1+t} + \frac{1}{1-t}\;dt$

$=\int \frac{1}{1+t}\;dt - \int \frac{-1}{1-t}\;dt$

$=\log(1+t) -\log(1-t) =\log\frac{1+t}{1-t}=\log(\frac{1+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})})$.

According to CAS (see Fig. 1),

Fig. 1

However, the two results are equivalent as a CAS-aided verification (see Fig. 2) confirms their difference is a constant (see Corollary 2 in “Sprint to FTC“).

Fig. 2

Exercise-1 According to CAS,

Show that it is equivalent to the result obtained in Example-1

Exercise-2 Try

$\int \frac{1}{\sin(x)+1}\;dx$

$\int \frac{1}{\sin(x)+\cos(x)}\;dx$

$\int \frac{1}{(2+\cos(x))\sin(x)}\;dx$

$\int \frac{1}{5+4\sin(x)}\;dx$

$\int \frac{1}{2\sin(x)-\cos(x)+5}\;dx$

and of course,

$\int \frac{1+\sin(x)}{\sin(x)(1+\cos(x))}\;dx$

# Deriving Lagrange's multiplier

We wish to consider a special type of optimization problem:

Find the maximum (or minimum) of a function $f(x,y)$ subject to the condition $g(x,y)=0\quad\quad(1)$

If it is possible to solve $g(x)=0$ for $y$ so that it is expressed explicitly as $y=\psi(x)$, by substituting $y$ in (1), it becomes

Find the maximum (or minimum) of a single variable function $f(x, \psi(x))$.

In the case that $y$ can not be obtained from solving $g(x,y)=0$, we re-state the problem as:

Find the maximum (or minimum) of a single variable function $z=f(x,y)$ where $y$ is a function of $x$, implicitly defined by $g(x, y)=0\quad\quad\quad(2)$

Following the traditional procedure of finding the maximum (or minimum) of a single variable function, we differentiate $z$ with respect to $x$:

$\frac{dz}{dx} = f_x(x,y) + f_y(x,y)\cdot \frac{dy}{dx}\quad\quad\quad(3)$

Similarly,

$g_x(x,y) + g_y(x,y)\cdot \frac{dy}{dx}=0\quad\quad\quad(4)$

By grouping $g(x, y)=0$ and (3), we have

$\begin{cases} \frac{dz}{dx}= f_x(x, y)+g_x(x, y)\cdot \frac{dy}{dx}\\ g(x,y) = 0\end{cases}\quad\quad\quad(5)$

The fact that $\frac{dz}{dx}= 0$ at any stationary point $x^*$ means for all $(x^*, y^*)$ where $g(x^*, y^*)=0$,

$\begin{cases} f_x(x^*, y^*)+g_x(x^*, y^*)\cdot \frac{dy}{dx}\vert_{x=x^*}=0 \\ g(x^*,y^*) = 0\end{cases}\quad\quad\quad(6)$

If $g_y(x^*,y^*) \ne 0$ then from (4),

$\frac{dy}{dx}\vert_{x=x^*} = \frac{-g_x(x^*, y^*)}{g_y(x^*, y^*)}$

Substitute it into (6),

$\begin{cases} f_x(x^*, y^*)+f_y(x^*, y^*)\cdot (\frac{-g_x(x^*, y^*)}{g_y(x^*, y^*)})=f_x(x^*, y^*)+g_x(x^*, y^*)\cdot (\frac{-f_y(x^*, y^*)}{g_y(x^*, y^*)})\\ g(x^*,y^*) = 0\end{cases}\quad\quad\quad(7)$

Let $\lambda = \frac{-f_y(x^*, y^*)}{g_y(x^*, y^*)}$, we have

$f_y(x^*, y^*) + \lambda g_y(x^*, y^*) =0\quad\quad\quad(8)$

Combining (7) and (8) gives

$\begin{cases} f_x(x^*, y^*)+\lambda g_x(x^*, y^*) = 0 \\ f_y(x^*, y^*)+\lambda g_y(x^*, y^*)=0 \\ g(x^*, y^*) = 0\end{cases}$

It follows that to find the stionary points of $z$, we solve

$\begin{cases} f_x(x, y)+\lambda g_x(x, y) = 0 \\ f_y(x, y)+\lambda g_y(x, y)=0 \\ g(x, y) = 0\end{cases}\quad\quad\quad(9)$

for $x, y$ and $\lambda$.

This is known as the method of Lagrange’s multiplier.

Let $F(x,y,\lambda) = f(x,y) + \lambda g(x,y)$.

Since

$F_x(x,y,\lambda) = f_x(x,y) + \lambda g_x(x,y)$,

$F_y(x,y,\lambda)=f_y(x,y) + \lambda g_y(x,y)$,

$F_{\lambda}(x,y,\lambda) = g(x, y)$,

(9) is equivalent to

$\begin{cases} F_x(x, y, \lambda)=0 \\ F_y(x,y,\lambda)=0 \\ F_{\lambda}(x, y) = 0\end{cases}\quad\quad\quad(10)$

Let’s look at some examples.

Example-1 Find the minimum of $f(x, y) = x^2+y^2$ subject to the condition that $x+y=4$

Let $F(x, y, \lambda) = x^2+y^2+\lambda(x+y-4)$.

Solving

$\begin{cases}F_x=2x-\lambda=0 \\ F_y = 2y-\lambda = 0 \\ F_{\lambda} = x+y-4=0\end{cases}$

for $x, y, \lambda$ gives $x=y=2, \lambda=4$.

When $x=2, y=2, x^2+y^2=2^2+2^2=8$.

$\forall (x, y) \ne (2, 2), x+y=4$, we have

$(x-2)^2 + (y-2)^2 > 0$.

That is,

$x^2-4x+4 + y^2-4y+4 = x^2+y^2-4(x+y)+8 \overset{x+y=4}{=} x^2+y^2-16+8>0$.

Hence,

$x^2+y^2>8, (x,y) \ne (2,2)$.

The target function $x^2+y^2$ with constraint $x+y=4$ indeed attains its global minimum at $(x, y) = (2, 2)$.

I first encountered this problem during junior high school and solved it:

$(x-y)^2 \ge 0 \implies x^2+y^2 \ge 2xy$

$x+y=4\implies (x+y)^2=16\implies x^2+2xy +y^2=16\implies 2xy=16-(x^2+y^2)$

$x^2+y^2 \ge 16-(x^2+y^2) \implies x^2+y^2 \ge 8\implies z_{min} = 8$.

I solved it again in high school when quadratic equation is discussed:

$x+y=4 \implies y =4-x$

$z=x^2+y^2 \implies z = x^2+(4-x)^2 \implies 2x^2-8x+16-z=0$

$\Delta = 64-4 \cdot 2\cdot (16-z) \ge 0 \implies z \ge 8\implies z_{min} = 8$

In my freshman calculus class, I solved it yet again:

$x+y=4 \implies y=4-x$

$z = x^2+(4-x)^2$

$\frac{dz}{dx} = 2x+2(4-x)(-1)=2x-8+2x=4x-8$

$\frac{dz}{dx} =0 \implies x=2$

$\frac{d^2 z}{dx^2} = 4 > 0 \implies x=2, z_{min}=2^2+(4-2)^2=8$

Example-2 Find the shortest distance from the point $(1,0)$ to the parabola $y^2=4x$.

We minimize $f = (x-1)^2+y^2$ where $y^2=4x$.

If we eliminate $y^2$ in $f$, then $f = (x-1)^2+4x$. Solving $\frac{df}{dx} = 2x+2=0$ gives $x=-1$, Clearly, this is not valid for it would suggest that $y^2=-4$ from $y^2=4x$, an absurdity.

By Lagrange’s method, we solve

$\begin{cases} 2(x-1)-4\lambda=0 \\2y\lambda+2y = 0 \\y^2-4x=0\end{cases}$

Fig. 1

The only valid solution is $x=0, y=0, k=-\frac{1}{2}$. At $(x, y) = (0, 0), f=(0-1)^2+0^2=1$. It is the global minimum:

$\forall (x, y) \ne (0, 0), y^2=4x \implies x>0$.

$(x-1)^2+y^2 \overset{y^2=4x}{=}(x-1)^2+4x=x^2-2x+1+4x=x^2+2x+1\overset{x>0}{>}1=f(0,0)$

Example-3 Find the shortest distance from the point $(a, b)$ to the line $Ax+By+C=0$.

We want find a point $(x_0, y_0)$ on the line $Ax+By+C=0$ so that the distance between $(a, b)$ and $(x_0, y_0)$ is minimal.

To this end, we minimize $(x_0-a)^2+(y_0-b)^2$ where $Ax_0+By_0+C=0$ (see Fig. 2)

Fig. 2

We found that

$x_0=\frac{aB^2-bAB-AC}{A^2+B^2}, y_0=\frac{bA^2-aAB-BC}{A^2+B^2}$

and the distance between $(a, b)$ and $(x_0, y_0)$ is

$\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}\quad\quad\quad(11)$

To show that (11) is the minimal distance, $\forall (x, y) \ne (x_0, y_0), Ax+By+C=0$.

Let $d_1 = x-x_0, d_2=y-y_0$, we have

$x = x_0 + d_1, y=y_0 + d_2, d_1 \ne 0, d_2 \ne 0$.

Since $Ax+By+C=0$,

$A(x_0+d_1)+B(y_0+d_2)+C=Ax_0+Ad_1+By_0+Bd_2+C=0$

That is

$Ax_0+By_0+C+Ad_1+Bd_2=0$.

By the fact that $Ax_0+By_0+C=0$, we have

$Ad_1 + Bd_2 =0\quad\quad\quad(12)$

Compute $(x-a)^2+(y-b)^2 - ((x_0-a)^2+(y_0-b)^2)$ (see Fig. 3)

Fig. 3

yields

$\boxed{-\frac{2d_2BC}{B^2+A^2}-\frac{2d_1AC}{B^2+A^2}}+[\frac{d_2^2B^2}{B^2+A^2}]-\underline{\frac{2bd_2B^2}{B^2+A^2}}+(\frac{d_1^2B^2}{B^2+A^2})-\frac{2ad_2AB}{B^2+A^2}-\underline{\frac{2bd_1AB}{B^2+A^2}}+[\frac{d_2A^2}{B^2+A^2}]+(\frac{d_1^2A^2}{B^2+A^2})-\frac{2ad_1A^2}{B^2+A^2}$

After some rearrangement and factoring, it becomes

$\frac{-2C}{A^2+B^2}(Ad_1+Bd_2)+\frac{-2B}{A^2+B^2}(Ad_1+Bd_2)+\frac{-2A}{A^2+b^2}(Ad_1+Bd_2) + d_1^2+d_2^2$

By (12), it reduces to

$d_1^2 + d_2^2$.

This is clearly a positive quantity. Therefore,

$\forall (x, y) \ne (x_0, y_0), Ax+By+C=0 \implies (x-a)^2+(y-b)^2> (x_0-1)^2+(y_0-b)^2$

# Introducing Operator Delta

The $r^{th}$ order finite difference of function$f(x)$ is defined by

$\Delta^r f(x) = \begin{cases} f(x+1)-f(x), r=1\\ \Delta(\Delta^{r-1}f(x)), r > 1\end{cases}$

From this definition, we have

$\Delta f(x) = \Delta^1 f(x) = f(x+1)-f(x)$

and,

$\Delta^2 f(x) = \Delta (\Delta^{2-1} f(x))$

$= \Delta (\Delta f(x))$

$= \Delta( f(x+1)-f(x))$

$= (f(x+2)-f(x+1)) - (f(x+1)-f(x))$

$= f(x+2)-2f(x)+f(x+1)$

as well as

$\Delta^3 f(x) = \Delta (\Delta^2 f(x))$

$= \Delta (f(x+2)-2f(x)+f(x+1))$

$= (f(x+3)-2f(x+1)+f(x+2)) - (f(x+2)-2f(x)+f(x+1))$

$= f(x+3)-3f(x+2)+3f(x+1)-f(x)$

The function shown below generates $\Delta^r f(x), r:1\rightarrow 5$ (see Fig. 1).

delta_(g, n) := block(
local(f),

define(f[1](x),
g(x+1)-g(x)),

for i : 2 thru n do (
define(f[i](x),
f[i-1](x+1)-f[i-1](x))
),

return(f[n])
);


Fig. 1

Compare to the result of expanding $(f(x)-1)^r=\sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x)^{r-i}, r:1\rightarrow 5$ (see Fig. 2)

Fig. 2

It seems that

$\Delta^r f(x) = \sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x+r-i)\quad\quad\quad(1)$

Lets prove it!

We have already shown that (1) is true for $r= 1, 2, 3$.

Assuming (1) is true when $r=k-1 \ge 4$:

$\Delta^{k-1} f(x) = \sum\limits_{i=0}^{k-1}(-1)^i \binom{r}{i} f(x+k-1-i)\quad\quad\quad(2)$

When $r=k$,

$\Delta^k f(x) = \Delta(\Delta^{k-1} f(x))$

$\overset{(2)}{=}\Delta (\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i))$

$=\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+1+k-1-i)-\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

$=(-1)^0 \binom{k-1}{0}f(x+k-0)$

$+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

$-(-1)^{k-1}\binom{k-1}{k-1}f(x+k-1-(k-1))$

$\overset{\binom{k-1}{0} = \binom{k-1}{k-1}=1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$=f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)+\sum\limits_{i=0}^{k-2}(-1)^{i+1}\binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$\overset{j=i+1, i:0 \rightarrow k-2\implies j:1 \rightarrow k-1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{j=1}^{k-1}(-1)^j \binom{k-1}{j-1}f(x+k-j)-(-1)^{k-1}f(x)$

$= f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i-1}f(x+k-i)+(-1)^k f(x)$

$= f(x+k) + \sum\limits_{i=1}^{k-1}(-1)^i f(x+k-i) (\binom{k-1}{i} + \binom{k-1}{i-1})+(-1)^k f(x)$

$\overset{\binom{k-1}{i} + \binom{k-1}{i-1}=\binom{k}{i}}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k f(x)$

$= (-1)^0 \binom{k}{0}f(x+k-0)+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k \binom{k}{k} f(x+k-k)$

$= \sum\limits_{i=0}^{k}(-1)^i \binom{k}{i} f(x+k-i)$

# A pair of non-identical twins

A complex number $x + i y$ can be plotted in a complex plain where the $x$ coordinate is the real axis and the $y$ coordinate the imaginary.

Let’s consider the following iteration:

$z_{n+1} = z_{n}^2 + c\quad\quad\quad(1)$

where $z, c$ are complex numbers.

If (1) are started at $z_0 = 0$ for various values of $c$ and plotted in c-space, we have the Mandelbrot set:

When $c$ is held fixed and points generated by (1) are plotted in z-space, the result is the Julia set:

# 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 $y=kx+m$:

$x=-r, x=r$