Family of lines that are at a distance of 5 from the origin

[vcsharp2003]

Homework Statement

Write the equation of the family of lines that are at a distance of 5 from the origin.

Relevant Equations

y = mx + c, where m is the slope of the straight line and c is its y-intercept

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached.

The one-line solution in the book says: The equation is $x \cos{\omega} +y \sin{\omega} - 5 = 0$, $\omega$ being the parameter.

From my side, the only thing I could come up with is that if we consider a circle centered at (0,0) and having a radius of 5, then all the tangent lines to such a circle would be the family of lines being asked in this question. I do not know any formula or method to specify this family of tangents.

My question is, how did the book arrive at the one-line solution?

 

[anuttarasammyak]

The nearest point of the lines are on the circle x^2+y^2=5^2. The lines are orthogonal to the circle radius.

 

[vcsharp2003]

The nearest point of the lines are on the circle x^2+y^2=5^2. The lines are orthogonal to the circle radius.

I get that since a tangent to a circle will always be perpendicular to its radius. But, the question still remains how the one-liner solution was arrived at and what does $\omega$ variable in the solution represent? Maybe it involves a long derivation.

 

[fresh_42]

The straight line through the origin and a point on the circle has the slope $\dfrac{\sin \omega}{\cos \omega}.$ This means that the tangent has the slope $-\dfrac{\cos \omega}{\sin \omega}.$ Hence, the tangent line writes $y=-\dfrac{\cos \omega}{\sin \omega} \cdot x +b$ and we must determine $b,$ which is the distance on the $y$ axis between the origin and the intersection of the tangent with the $y$ axis.

This needs a bit of geometry on a right triangle. It will turn out to be $b=\dfrac{5}{\sin \omega}$ using the definition of the sine as opposite / hypotenuse.

 

[anuttarasammyak]

I get that since a tangent to a circle will always be perpendicular to its radius. But, the question still remains how the one-liner solution was arrived at and what does $\omega$ variable in the solution represent? Maybe it involves a long derivation.

$$y-5\sin\omega=-\frac{1}{\tan\omega}(x-5\cos\omega)$$

[EDIT]
Another way. Say all the points on the line
$$ y=ax+b $$
has distance more than c from the origin.
$$ x^2 + y^2 \geq c^2 $$
Deleting y
$$ (1+a^2)(x+\frac{ab}{1+a^2})^2+[b^2-c^2-\frac{a^2b^2}{1+a^2}] \geq 0 $$
[ ] must be zero or positive. The marginal case zero
$$ b^2=c^2(1+a^2)$$
Thus the line is with real number parameter a
$$ y = ax　\pm c \sqrt{1+a^2} $$

 

[pasmith]

The equation of a line in $\mathbb{R}^2$ can be written as $$\mathbf{n} \cdot (\mathbf{x} - \mathbf{x}_0) = 0$$ where $\mathbf{x}_0$ is a point on the line and $\mathbf{n}$ is a vector normal to the line. Here we have $\mathbf{n} = (\cos \omega, \sin \omega)$ and $\mathbf{x}_0 = (5\cos \omega, 5 \sin \omega)$. The result follows.

 

[fresh_42]

The equation of a line in $\mathbb{R}^2$ can be written as $$\mathbf{n} \cdot (\mathbf{x} - \mathbf{x}_0) = 0$$ where $\mathbf{x}_0$ is a point on the line and $\mathbf{n}$ is a vector normal to the line. Here we have $\mathbf{n} = (\cos \omega, \sin \omega)$ and $\mathbf{x}_0 = (5\cos \omega, 5 \sin \omega)$. The result follows.

Precalculus?

 

[vcsharp2003]

I am trying to understand the variable $\omega$, since the one-liner solution makes no mention of it.
It seems that this variable is the angle from the positive x-axis to the radius in an anti-clockwise direction, as shown in the diagram below. If that's correct, then the solution should be as below unless this variable means something else.

By using the tangent circle diagram below, I could find the x and y intercepts, which are a and b respectively.
Then I could use the intercept form equation of a straight line, which is $\frac{x}{a} + \frac{y}{b} = 1$ to get the general equation of any tangent to the given circle of radius 5 centered at (0,0).

From the two right-angle triangles shown in the diagram above, $\cos {\omega} = \frac{5}{a}$ and $\sin {\omega} = \frac{5}{b}$.
$$\therefore a = \frac{5}{\cos{\omega}}$$
$$\text{and } b = \frac{5}{\sin{\omega}}$$
Substituting the above in the intercept form of the equation, we get
$$ \frac{x}{\frac{5}{\cos{\omega}}} + \frac{y}{\frac{5}{\sin{\omega}}} = 1$$
$$\therefore {x} {\cos{\omega}} + {y} {\sin{\omega}} = 5 $$
$$\implies {x} {\cos{\omega}} + {y} {\sin{\omega}} - 5 = 0 $$

The above equation is what's given in the one-liner solution, and it represents a family of lines that are at a distance of 5 from the origin, with a single parameter of $\omega$, which is the angle made by the perpendicular from the origin to a straight line belonging to this family with the positive x-axis in radians ( this angle must be taken in an anti-clockwise direction).

Each real value of $\omega$, whether positive or negative or zero, will give us the equation of a straight line belonging to this family.

 

[fresh_42]

That's correct.

Theoretically, you have to distinguish between the cases where $\sin \omega=0$ and those where $\sin \omega\neq 0$ because we divided by $\sin \omega.$ This can be resolved by taking vectors instead of slopes. Here we avoided it by multiplying the equation with $\sin \omega,$ but that is a bit imprecise since it is not clear that we can do this: multiplication by $0$ to avoid division by $0$ is a bit strange. However, if we use slopes, then two lines of the bundle have an infinite slope, and two lines a zero slope. Vectors consider both coordinates equally as components, and it doesn't matter if they are zero or not, because no division occurs. So, it makes a difference whether you can use them as in post #6 or you have to use slopes as I suggested in post #4.

The vector notation considers
$$
\begin{pmatrix}x\\y\end{pmatrix}\cdot \begin{pmatrix}v_x\\v_y\end{pmatrix}= c
$$
and the slope notation
$$
y=mx+b
$$
which requires a special treatment for $x=a.$

 

[vcsharp2003]

That's correct.

Theoretically, you have to distinguish between the cases where $\sin \omega=0$ and those where $\sin \omega\neq 0$ because we divided by $\sin \omega.$ This can be resolved by taking vectors instead of slopes. Here we avoided it by multiplying the equation with $\sin \omega,$ but that is a bit imprecise since it is not clear that we can do this: multiplication by $0$ to avoid division by $0$ is a bit strange. However, if we use slopes, then two lines of the bundle have an infinite slope, and two lines a zero slope. Vectors consider both coordinates equally as components, and it doesn't matter if they are zero or not, because no division occurs. So, it makes a difference whether you can use them as in post #6 or you have to use slopes as I suggested in post #4.

The vector notation considers
$$
\begin{pmatrix}x\\y\end{pmatrix}\cdot \begin{pmatrix}v_x\\v_y\end{pmatrix}= c
$$
and the slope notation
$$
y=mx+b
$$
which requires a special treatment for $x=a.$

So, I must deal with the special cases $\sin{\omega} =0$ and $\cos{\omega} =0$ in my proposed solution.

 

[fresh_42]

So, I must deal with the special cases $\sin{\omega} =0$ and #\cos{\omega} =0## in my proposed solution.

I think you don't need $a,$ so you have only to deal with $\sin \omega=0$ or $\omega\in \{0,\pi\}$ since we may assume that $\omega\in [0,2\pi).$ These are the vertical lines, i.e. the lines $x=\pm 5.$

So it is sufficient to prove the formula for $\sin \omega\neq 0$ and show that the formula also delivers the two remaining cases $y=\pm 5$ when $\sin \omega=0.$

If you continue using $a$, then you have four cases to exclude, and you must show that the resulting formula without quotients captures these four cases, too.

Or you use vectors, which treat both components equally.

 

[vcsharp2003]

It seems complex, but I will try. So, I am going to restrict to the interval $\omega \in [0,2\pi)$ since all other values of $\omega$ will simply be a repeat of the corresponding value in this interval.

One thing that comes to my mind is that the edge cases would be $x = \pm {5}$ and $y = \pm {5}$, which are tangents to the considered circle and therefore a part of the family of lines.
So, we could say that the family of lines that are a distance of 5 from origin is represented by the following two sets of lines:

1. ${x} {\cos{\omega}} + {y} {\sin{\omega}} - 5 = 0$ where ${\omega} \not\in \{0, \frac {\pi}{2}, \pi, \frac {3\pi}{2} \}$ (my post# 8 gives this part of the solution)

2. the four lines represented by $x = \pm {5}$ for $\omega \in \{0,\pi\}$ and $y = \pm {5}$ for $\omega \in \{\frac{\pi}{2},\frac{3\pi}{2}\}$

Then by inspection of the first set of lines, we see that substituting $0$ or $\pi$ or $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ in the equation ${x} {\cos{\omega}} + {y} {\sin{\omega}} - 5 = 0$ gives us the lines $x = \pm {5}$ and $y = \pm {5}$.
$\therefore$ we can say that the family of lines can be represented by a single equation ${x} {\cos{\omega}} + {y} {\sin{\omega}} - 5 = 0$ where ${\omega} \in [0,2\pi\}$

Would the above reasoning be acceptable?

 

[fresh_42]

It seems complex, but I will try. So, I am going to restrict to the interval $\omega \in [0,2\pi)$ since all other values of $\omega$ will simply be a repeat of the corresponding value in this interval.

One thing that comes to my mind is that the edge cases would be $x = \pm {5}$ and $y = \pm {5}$, which are tangents to the considered circle and therefore a part of the family of lines.
So, we could say that the family of lines that are a distance of 5 from origin is represented by the following two sets of lines:

1. ${x} {\cos{\omega}} + {y} {\sin{\omega}} - 5 = 0$ where ${\omega} \not\in \{0, \frac {\pi}{2}, \pi, \frac {3\pi}{2} \}$ (my post# 8 gives this part of the solution)

2. the four lines represented by $x = \pm {5}$ for $\omega \in \{0,\pi\}$ and $y = \pm {5}$ for $\omega \in \{\frac{\pi}{2},\frac{3\pi}{2}\}$

Then by inspection of the first set of lines, we see that substituting $0$ or $\pi$ or $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ in the equation ${x} {\cos{\omega}} + {y} {\sin{\omega}} - 5 = 0$ gives us the lines $x = \pm {5}$ and $y = \pm {5}$.
$\therefore$ we can say that the family of lines can be represented by a single equation ${x} {\cos{\omega}} + {y} {\sin{\omega}} - 5 = 0$ where ${\omega} \in [0,2\pi\}$

Would the above reasoning be acceptable?

Yes.

My only remark is that you should call this "family" a bundle or, more specifically, a tangent bundle. This is the correct technical term. It doesn't matter in this case, but why not do it right in the first place?

 

[vcsharp2003]

Yes.

My only remark is that you should call this "family" a bundle or, more specifically, a tangent bundle. This is the correct technical term. It doesn't matter in this case, but why not do it right in the first place?

Thank you for helping me solve this problem..

 

[vcsharp2003]

The equation of a line in $\mathbb{R}^2$ can be written as $$\mathbf{n} \cdot (\mathbf{x} - \mathbf{x}_0) = 0$$ where $\mathbf{x}_0$ is a point on the line and $\mathbf{n}$ is a vector normal to the line. Here we have $\mathbf{n} = (\cos \omega, \sin \omega)$ and $\mathbf{x}_0 = (5\cos \omega, 5 \sin \omega)$. The result follows.

Is n a unit vector? And do x and x₀ also denote vectors?

My guess is that n can denote a normal unit vector or a normal non-unit vector, but it must be normal to the straight line whose equation is being sought. And x denotes the position vector of any variable point on the straight line whose equation we are trying to find, while x₀ denotes the position vector of a known point on the same straight line.

Am I correct?

The vectors n and x₀ must be known to use this formula for a straight line.

We could rewrite this formula as below, if we are in the x-y plane.

$$\hat {n} \cdot (x\hat{i} + y\hat{j} - x_0\hat{i} + y_0\hat{j}) = 0$$

OR the form below

$$\hat {n} \cdot ((x-x_0)\hat{i} + (y-y_0)\hat{j}) = 0$$


Also, a unit vector along any straight line in the x-y plane would be $\hat {n} = \cos {\theta} \hat {i} + \sin{\theta} \hat {j}$ where $\theta$ is the smallest angle in a counter clockwise direction made by the vector $\hat {n}$ with the positive x axis.

We could further rewrite this formula as follows, in which we need to know the angle $\theta$ and a point $(x_0,y_0)$ on the straight line.

$$(\cos {\theta} \hat {i} + \sin{\theta} \hat {j}) \cdot ((x-x_0)\hat{i} + (y-y_0)\hat{j}) = 0$$