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.

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