Circles tangential to the real line

Question

For any two distinct points $a,b$ in the upper half plane, prove that there is a circle passing through $a$ and $b$ that is tangential to the extended real line.

Seems obvious, but how would you go about proving this?

Answer 1

This is a fascinating question.

Here’s how I would argue, using the group of rigid motions of the Poincaré (hyperbolic) plane.

What are the (general) circles tangent to the (extended) real line at $\infty$? They are the horizontal lines $Y=c$, for positive real $c$. I suggest applying a rotation $\rho$ about $a=(x_a,y_a)$ in the hyperbolic plane that brings $b$ to a point $b'$ whose height above the real axis is equal to that of $a$, so that $b'=(x_{b'},y_a)$. Now draw the horizontal line $Y=y_a$, and apply $\rho^{-1}$ to this picture. What happens is that $b'$ rotates back to $b$, $\infty$ rotates (most likely) to a finite point $p=(x_p,0)$ on the real axis, and our horizontal line rotates to the circle through $a$, $p$, and $b$, tangent to the real axis at $p$.

Answer 2

I'm going to consider the case where both points lie on the real line as having been addressed in the comments. So I can assume that $a$ is not on the real line.

The set of points equidistant from the real line and $a$ forms a parabola, $P$, in the upper half plane. The circle center $P$ must lie on this.

The set of points equidistant from $a$ and $b$ forms a line. $P$ must also lie on this. So $P$ is at the intersection of a line and a parabola, hence is one of at most two points. That seems like a pretty fast route, since it involves solving exactly one quadratic.

If the point $a$ is given by $(2r, 2s)$, then the equation of the parabola is $$ (x-2r)^2 + (2s)^2 = 4sy $$ If the point $b = (2p, 2q)$, then the midpoint of $ab$ is $(p+r, q+s)$, and the parametric equation of the perpendicular bisector of $ab$ is $$ x(t) = (p+r) + t (s-q) \\ y(t) = (q + s) + t(p-r) $$ Plugging this in, you get

$$ [p+r + t(s-q) - 2r]^2 + 4s^2 = 4s [q + s + t(p-r)] $$ That's a quadratic in $t$, which you can solve with the quadratic equation, and then plug the result back into the line equation to get the two circle centers.

Doing the algebra: \begin{align} [p-r + t(s-q)]^2 + 4s^2 &= 4s [q + s + t(p-r)]\\ (p-r)^2 + 2t(s-q)(p-r) + t^2 (s-q)^2 &= - 4s^2 +4s(q + s) + t\cdot 4s(p-r)\\ (s-q)^2 t^2 + 2t(s-q)(p-r) -2t\cdot 2s(p-r)+(p-r)^2 &= - 4s^2 +4s(q + s)\\ (s-q)^2 t^2 + 2t(p-r)[(s-q) -2s] +(p-r)^2 &= - 4s^2 +4s(q + s)\\ (s-q)^2 t^2 + 2t(p-r)[-(s+q)] +(p-r)^2 &= - 4s^2 +4s(q + s)\\ (s-q)^2 t^2 - 2t(p-r)(s+q) + (p-r)^2 - 4sq &= 0\\ (s-q)^2 t^2 - 2t(p-r)(s+q) + (p-r)^2 - 4sq &= 0\\ (s-q)^2 t^2 - 2t(p-r)(s+q) + (p-r)^2 - 4sq &= 0 \end{align} That sure looks as if there ought to be more $s-q$ terms, and not $s+q$ or $4sq$, but who knows? Anyhow, it's a general guide to how to solve this.

Actually, on reflection, it looks pretty good to me. For if the points $a$ and $b$ have the same $y$-coordinate, then $s-q = 0$, and the quadratic becomes linear...which makes some sense to me --- I can only visualize one circle that passes through both points and is tangent to the $x$-axis. On the other hand, if the $x$-coordinates are the same, then the coefficient of $t$ becomes $0$, and the two roots are negatives of one another, corresponding to the reflection (about a vertical axis) symmetry in the situation. So: except in the $s = q$ case, we get \begin{align} t &= \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\\ &= \frac{2(p-r)(s+q) \pm \sqrt{4(p-r)^2(s+q)^2 + 4(s-q)^2 4sq}}{2(s-q)^2}\\ &= \frac{(p-r)(s+q) \pm \sqrt{(p-r)^2(s+q)^2 + 4(s-q)^2 sq}}{(s-q)^2}\\\end{align} and then, with these $t$-values, we have $$ x = (p+r) + t (s-q) \\ y = (q + s) + t(p-r) $$

Answer 3

Write $A=a$ and $B=b$.

Suppose line $AB$ cuts real line $\ell$ at $P$ and draw arbitrary circle $C$ through $A$ and $B$. Now draw a tangent from $P$ to $C$ and let it touch $C$ at $E$. Then draw a circle with center at $P$ and radius $PE$. This circle cuts $\ell$ at $D_1$ and $D_2$. Now let perpendicular bisector of $AB$ cuts line $d_i$ which is perpendicular to $\ell$ and goes through $D_i$ at $S_i$. You get the center(s) of circle(s) you seek for.

Proof that this construction is OK. Observe the power of the point of $P$ with respect to the given circles ($P$ is radical center).