More on the circular solution to the intersection of two perpendicular lines

In the last post, I showed that the intersection of two perpendicular lines must lie on a circle, so long as the lines are each forced to go through particular points. The final result was a parameterization based on the classic cosine, sine version of a circle, but the bit I found more interesting was the earlier form:

(x,y) = \left(\cfrac{am^2+(d-c)m+b}{m^2+1},\cfrac{dm^2+(b-a)m+c}{m^2+1}\right), \enspace m\in \mathbb{R}

One of the results of the parameterization was that the angle at which the point lied on the circle was not the angle at which one of the lines made with the x axis (unless the circle’s center was at the origin). This led to a phase shift in the parameterization from the angle of the line. If we were willing to lose a bit of information, the phase, we could also show that the above is a circle if it satisfies

\left(x-\cfrac{a+b}{2}\right)^2+\left(y-\cfrac{c+d}{2}\right)^2 = R^2

with

R^2 = \left(\cfrac{b-a}{2}\right)^2 + \left(\cfrac{d-c}{2}\right)^2

Since we already have the parameterization with above, showing this is true is just a matter of algebra. To start, add and subtract the coordinates of the center of the circle from x and y:

\begin{array}{c} x = \cfrac{am^2+(d-c)m+b}{m^2+1}-\cfrac{a+b}{2}+\cfrac{a+b}{2}\\\\ y=\cfrac{dm^2+(b-a)m+c}{m^2+1}-\cfrac{c+d}{2}+\cfrac{c+d}{2}\end{array}

Finding a common denominator and carefully combining gives

\begin{array}{c} x = \cfrac{(a-b)m^2+2(d-c)m+b-a}{2m^2+2}+\cfrac{a+b}{2}\\\\ y=\cfrac{(d-c)m^2+2(b-a)m+c-d}{2m^2+2}+\cfrac{c+d}{2}\end{array}

We now have a form that, when plugged into the LHS for the circle equation, cancels out the center point coordinates.

\left(x-\cfrac{a+b}{2}\right)^2+\left(y-\cfrac{c+d}{2}\right)^2 = \left(\cfrac{(a-b)m^2+2(d-c)m+(b-a)}{2m^2+2}\right)^2+\left(\cfrac{(d-c)m^2+2(b-a)m+(c-d)}{2m^2+2}\right)^2

The RHS of this thing is a bit easier to work with with by letting

p = b-a \quad\text{and}\quad q=d-c.

It becomes

\left(\cfrac{-pm^2+2qm+p}{2m^2+2}\right)^2+\left(\cfrac{qm^2+2pm-q}{2m^2+2}\right)^2

Careful manipulation yields

(p^2+q^2)\cfrac{m^4+2m^2+1}{2m^2+2}

Nicely, the m‘s cancel out completely. This becomes

\cfrac{p^2+q^2}{4} = \left(\cfrac{p}{2}\right)^2+\left(\cfrac{q}{2}\right)^2 = \left(\cfrac{b-a}{2}\right)^2+\left(\cfrac{d-c}{2}\right)^2

Done!

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