find the two circles inscribed by three lines

Question

Let $l$ be the (non vertical) line with equation $y=mx+b$. $l$ determines two circles tangent to $l$ and the two vertical lines: $x=-1$, $x=1$. Clearly, the centres of these circles both lie on the y-axis. What are their y-coordinates?

Answer 1

I just realized that, viewed algebraically, the solution is easy:

The equation of the circles are $x^2 + (y-y_0)^2 = 1$ and the line is $y=mx+b$. Intersection is given by a quadratic equation in $x$ (with coefficients in terms of $y_0$). The tangency condition (algebraically, set the discriminant to 0) gives a quadratic in $y_0$.

Answer 2

Here is a visual representation for finding the y-coordinate for the circle above the tangent line $y=mx+b$:

The height of the circle is merely the green line plus the orange line. Now how do we find these?

The green line isn't too hard to set up. The height is just $mx+b$ for the point that is tangent to the circle. We will get back to this in a second...

Finding the orange line is a tad harder, so let's use some trigonometry. Notice that we are dealing with the unit circle here, so the length of the orange line would be $|\sin{\theta}|$.

We actually can find $\theta$ (the angle made between a line from the center of the circle to the tangent point with the horizontal axis). First, see that for the circle, the angle $\theta$ being made is perpendicular to the line $mx+b$. The angle the line makes with the x-axis is equal to $\tan^{-1} (m)$. The line perpendicular to $mx+b$ has a slope of $-1/m$, so we now have $\theta = \tan^{-1}(-1/m)$

Thus, the orange line has a length of $$\color{#ffa500}{\left|\sin\left(\tan^{-1}\left(\frac{-1}{m}\right)\right)\right|}$$

Getting back to the green length. We need to find the $x$ value of the tangent point. This is merely $\cos(\theta)$.

Plugging our $x$ value into the equation of the line gives the green length: $$\color{green}{\left|m\left(\cos(\tan^{-1}\left(\frac{-1}{m}\right)\right)\right| + b}$$

(Yes, the absolute value is necessary, otherwise we run into problems when our line has negative slope)

The $y$ coordinate of the center of the circle is simply the sum of the two values we calculated.

A similar process can be done to find the coordinate for the bottom circle's coordinates.

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Edit: in the case of $m=0$, the y-coordinate is merely $1+b$

Answer 3

Given $m=\tan\phi$, $y(0)=b$, the centers $O_1,O_2$ are located at vertical distance $d$ from the point $C(0,b)$,

\begin{align} O_1&=(0,b+d) ,\\ O_2&=(0,b-d) ,\\ d&=\frac1{\cos\phi} =\sqrt{m^2+1} . \end{align}