Determining the curve to which variable line segments are tangent

Question

A variable line passing through a fixed point $P(x_0,y_0)$ intersects the circle $x^2+y^2=a^2$ at some point $Q$. If all the perpendicular bisectors of the variable line segment $PQ$ are tangent to a curve $C$, determine $C$ if (i) $P$ is inside the circle (ii) $P$ is outside the circle (Note: in both cases, only select one point of intersection with the circle, do not find the perpendicular bisector of the chord between both points of intersection);

Motivation:

I was watching this video on Feynmann's lost lecture on planetary orbits and came across an interesting geometry puzzle; https://youtu.be/xdIjYBtnvZU?si=g-5pJiu_zo8SZf5_ which moreover resolves the first part of the question but then my dense self could not understand why the center, $Q$, and $P$ as defined in "geometry proof land" are collinear so I decided to turn to a hardcore coordinate geometry approach to the problem. I would really appreciate it if the answer to this problem also uses the tools of coordinate geometry.

My attempt:

Let the variable line be $y-y_0=m(x-x_0)$. Substituting into the circle's equation, I got:

$(1+m^2)x^2+2m(y_0-mx_0)x+(mx_0-y_0)^2-a^2=0$.

And, $(1+m^2)y^2+2(mx_0-y_0)y+(mx_0-y_0)^2-m^2a^2=0$

I was thinking of perhaps explicitly finding the intersection point of the circle and variable line, say, some $(x,y)$, then writing the midpoint of $PQ$ as $(h,k)=\left(\dfrac{x+x_0}{2}, \dfrac{y+y_0}{2} \right)$, then eliminating $m$ and finding a relation between $h$ and $k$ which would then be the required locus.

But there are many problems with this approach; firstly, I'm implicitly assuming that the perpendicular bisector is tangent at the point it intersects $PQ$ to the curve $C$ without any justification. Secondly, the expressions for $x$ and $y$ are really large (so eliminating m becomes a very challenging task) and I don't know which sign I am supposed to take. Thirdly, I had a gut feeling that since the case considered in the video turns out to be an ellipse, the other case may either be a hyperbola or just a simple ellipse but I have no idea how to prove this intuition either.

Can anyone help me figure this out the coordinate geometry way?

Answer 1

Let me see if I can convince you with a different geometrical approach, instead of resorting to coordinates. I'll do my proof for a point $P$ inside the circle, but a similar reasoning is possible if $P$ is outside (see EDIT at the end).

Consider two rays through $P$, $PQ$ and $PQ'$, with their perpendicular bisectors meeting at $\tilde T$ (see figure below). Midpoints $M$ and $M'$ lie on a fixed circle (purple in the figure), which is the transformed of the given circle under a homothety of centre $P$ and ratio $1\over2$. Hence its centre $C$ is the midpoint of $OP$ and its radius ${1\over2}a$

But points $M$ and $M'$ also belong to the circle with diameter $P\tilde T$ (blue in the figure). If $Q'$ is moved towards $Q$, point $\tilde T$ also moves, approaching a limiting position $T$ which is the point on the envelope of the perpendicular bisectors lying on the perpendicular bisector of $PQ$ (this is the "classical" definition of envelope you can find on Wikipedia).

When $Q'\to Q$ also $M'\to M$ and circle $PM'M\tilde T$ tends then to a circle $PMT$ (see figure below) which is tangent (at $M$) to the circle of centre $C$. This fixes the centre $D$ of circle $PMT$ (because $D$ lies both on $CM$ and on the perpendicular bisector of $PM$) and also fixes $T$ because $PT$ is a diameter.

From $PD=DT$ and $PC=CO$ it follows $CM\parallel OT$. But from $PM=MQ$ it also follows $CM\parallel OQ$. Hence $OTQ$ are aligned and $T$ is the intersection between $OQ$ and the perpendicular bysector of $PQ$.

As a consequence we have $PT=QT$ and $PT+OT=QT+OT=a$. The locus of $T$ is thus the ellipse (dotted green in the figure) having foci $O$, $P$ and major axis $a$.

EDIT.

In the case when $P$ is outside the circle, one can repeat the same argument as above. I won't show all the details but just give below the analogous of the last figure. In this case $T$ lies outside $OQ$ and we have: $PT-OT=QT-OT=a$. The locus of $T$ is thus the hyperbola having foci $O$, $P$ and transverse axis $a$.

Answer 2

Your methood is perfectly fine. $C$ is an ellipse when $P$ is inside, a hyperbola when it's outside. The equation in both cases is

$$(4a^2-4x_0^2)x^2-8x_0y_0xy+(4a^2-4y_0^2)y^2+(-4a^2x_0+4x_0^3+4x_0y_0^2)x+(-4a^2y_0+4x_0^2y_0+4y_0^3)y-a^4+2a^2x_0^2-x_0^4+2a^2y_0^2-2x_0^2y_0^2-y_0^4=0$$

Found in M2 by

R=QQ[a,m,x0,y0][h,k,x,y,MonomialOrder=>Eliminate 2]
I=ideal(h^2+k^2-a^2,k-y0-m*(h-x0)) -- define Q = (h,k)
J=ideal((y-(k+y0)/2)*m+(x-(h+x0)/2))+I -- define the perpendicular bisector, it's of slope -1/m and goes through ((h+x0)/2,(k+y0)/2).
gens gb J -- this eliminates h,k
f=oo_0_0 -- 4*x^2+8*m*x*y+4*m^2*y^2+(-4*m*y0-4*x0)*x+(-4*m^2*y0-4*m*x0)*y-a^2*m^2+m^2*x0^2+m^2*y0^2-a^2+x0^2+y0^2 -- the line pair of parallel lines, comprising the two choices for Q
R=QQ[a,x0,y0][m,x,y,MonomialOrder=>Eliminate 1]
K=ideal(f, diff(m,f)) -- it so happens we don't need to disentangle them
gens gb K
g=oo_0_0 -- (4*a^2-4*x0^2)*x^2-8*x0*y0*x*y+(4*a^2-4*y0^2)*y^2+(-4*a^2*x0+4*x0^3+4*x0*y0^2)*x+(-4*a^2*y0+4*x0^2*y0+4*y0^3)*y-a^4+2*a^2*x0^2-x0^4+2*a^2*y0^2-2*x0^2*y0^2-y0^4

Discussing the equation I get $$(x-x_0)^2+(y-y_0)^2=e^2\frac{((x_0 x+ y_0 y-a^2)-(x_0^2+y_0^2-a^2)/2)^2}{x_0^2+y_0^2},$$ where $e$ is the eccentricity.

Answer 3

It relies on a known result (up to a homothety with ratio $1/2$ as explained in the answer of @intelligenci pauca) ; consider this excerpt from

https://mathshistory.st-andrews.ac.uk/Curves/Definitions2/

"Negative pedal: Given a curve C and O a fixed point then for a point P on C draw a line perpendicular to OP. The envelope of these lines as P describes the curve C is the negative pedal of C. The ellipse is the negative pedal of a circle if the fixed point is inside the circle while the negative pedal of a circle from a point outside is a hyperbola."