# The Missing Intercept

• March 29th 2009, 03:26 AM
Soroban
The Missing Intercept
The Missing Intercept

I posted this puzzle some time ago,
but no one provided a satisfactory answer.

Given: . $\begin{Bmatrix}x \:=\:\dfrac{1-t^2}{1+t^2} \\ \\[-3mm] y \:=\:\dfrac{2t}{1+t^2} \end{Bmatrix}$

We have the parametric equations of a unit circle.

Verification

Square: . $\begin{Bmatrix}x^2 \:=\:\dfrac{(1-t^2)^2}{(1+t^2)^2} & [1]\\ \\[-3mm] y^2 \:=\:\dfrac{(2t)^2}{(1+t^2)^2} & [2]\end{Bmatrix}$

Add [1] and [2]: . $x^2+y^2\:=\:\frac{1 - 2t^2 + t^4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2}$

$= \;\frac{1 + 2t^2+t^4}{(1+t^2)^2} \:=\:\frac{(1+t^2)^2}{(1+t^2)^2} \;=\;1$

Hence, a unit circle: . $x^2+y^2\:=\:1$

To find the $y$-intercepts, let $x = 0.$

$x = 0\!:\;\;\frac{1-t^2}{1+t^2}\:=\:0 \quad\Rightarrow\quad t \:=\:\pm1$

Hence, the $y$-intercepts are: . $(0,1),\;(0,\text{-}1)$

To find the $x$-intercepts, let $y = 0.$

$y = 0\!:\;\;\frac{2t}{1+t^2} \:=\:0 \quad\Rightarrow\quad t \:=\:0$

Hence, the $x$-intercept is: . $(1,0)$ ?

Where is the other $x$-intercept?

• March 29th 2009, 04:16 AM
running-gag
Hi

I don't know which answers have been given previously so I am trying (Happy)

Let E the set defined by $\begin{Bmatrix}x \:=\:\dfrac{1-t^2}{1+t^2} \\ \\[-3mm] y \:=\:\dfrac{2t}{1+t^2} \end{Bmatrix}$ for t real

By showing that $x^2+y^2=1$ you are proving that E is included in the unit circle, but not that it is equal to the unit circle. And it is not equal since the point (-1,0) is not inside E (there is no value of t such that x=-1 and y=0). Only if you are allowed to consider infinite values of t, you can find this point.
• March 29th 2009, 06:15 AM
CaptainBlack
Quote:

Originally Posted by Soroban
The Missing Intercept

I posted this puzzle some time ago,
but no one provided a satisfactory answer.

Given: . $\begin{Bmatrix}x \:=\:\dfrac{1-t^2}{1+t^2} \\ \\[-3mm] y \:=\:\dfrac{2t}{1+t^2} \end{Bmatrix}$

We have the parametric equations of a unit circle.

Verification

Square: . $\begin{Bmatrix}x^2 \:=\:\dfrac{(1-t^2)^2}{(1+t^2)^2} & [1]\\ \\[-3mm] y^2 \:=\:\dfrac{(2t)^2}{(1+t^2)^2} & [2]\end{Bmatrix}$

Add [1] and [2]: . $x^2+y^2\:=\:\frac{1 - 2t^2 + t^4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2}$

$= \;\frac{1 + 2t^2+t^4}{(1+t^2)^2} \:=\:\frac{(1+t^2)^2}{(1+t^2)^2} \;=\;1$

Hence, a unit circle: . $x^2+y^2\:=\:1$

To find the $y$-intercepts, let $x = 0.$

$x = 0\!:\;\;\frac{1-t^2}{1+t^2}\:=\:0 \quad\Rightarrow\quad t \:=\:\pm1$

Hence, the $y$-intercepts are: . $(0,1),\;(0,\text{-}1)$

To find the $x$-intercepts, let $y = 0.$

$y = 0\!:\;\;\frac{2t}{1+t^2} \:=\:0 \quad\Rightarrow\quad t \:=\:0$

Hence, the $x$-intercept is: . $(1,0)$ ?

Where is the other $x$-intercept?

Not much of a puzzle it's at $t=\pm\infty$ (or rather it's the limit point as $t \to \infty$ ) and there the point is $(-1,0)$. So the curve never reaches the second intercept and the curve is the unit circle with a hole at $(-1,0).$

CB
• March 29th 2009, 11:19 AM
Soroban

Thank you, running-gag and The Cap'n!

Those are the answers I was hoping for.