Three other (more traditional) proofs of circle of Apollonius

apollonius

We prove that if the locus of z is a circle, then the locus of w=(az+b)/(cz+d) is a circle (or line).

Another proof is here: these three are more traditional.

First, let’s establish that without loss of generality we can work this one out just for the case that z is a circle of centre 0 and radius 1.

If the locus of z is a circle, then |z−k|=r for some centre k and radius r.

Let z'=(z−k)/r, and the locus of z' is the circle of centre 0 and radius 1.

If we write w in terms of z', it will be the same sort of equation as in terms of z: w'=(a'z'+b')/(c'z'+d') for some a', b', c', d'.

So we solve the whole problem if we solve it for:

w=(az+b)/(cz+d) ⇒ z=(dw−b)/(cw+a)

|z|=1 ⇒ |−b+dw|=|a−cw|

⇒ the distance of w from b/d is |c|/|d| times the distance of w from a/c

Thus the problem comes down to knowing the locus of a point A whose distance from a point C is λ times its distance from a point B. We will prove it is a circle (and find what circle it is) if λ≠1 (it’s called the Circle of Apollonius). We will prove it is a straight line if λ=1.

apollonius1

triangle2

First proof uses no trig. CL:LB=CM:MB=λ=CA:BA. We will show that the locus of A is a circle by proving that the line MA bisects the angle CAB (cuts into two equal halves) and the line LA bisects the angle CAB'. It follows that the angle LAM is half of the whole line B'AB, and so it is a right angle. Therefore A is always on the circle with diameter LM.

The proof breaks down if λ=1. In that case L zooms off to infinity. The locus of A is all points which are equally distant from B and from C, i.e. a straight line halfway between B and C and perpendicular to BC.

So, to prove MA and LA bisect the angles if λ≠1.

Look at the smaller diagram with triangle CAB. BD is drawn parallel to AC.

∠BDM=∠MAC and ∠DBM=∠MCA (alternate angles)

So ΔBDM is similar to ΔMAC

AC:BD=CM:BM=λ

But AC:AB=λ too

So BD=AB

ΔDBA is isosceles

Therefore ∠BAD (which=∠BAM) = ∠BDA = ∠MAC

Therefore MA bisects ∠BAC

Same sort of proof that LA bisects ∠CAB'  ▇

Another proof uses the sine rule.

CM/sin(MAC)=CA/sin(AMC) … [1]

BM/sin(MAB)=BA/sin(AMB) … [2]

Since ∠AMC+∠AMB = a straight line, sin(AMC)=sin(AMB); and CM/BM=CA/BA=λ

Divide equation [1] by equation [2]

sin(MAC)=sin(MAB)

Therefore ∠MAC=∠MAB  ▇

That’s shorter, but it assumes you already know how to prove the sine rule. The longer proof is nicer because it assumes less.

Or it can be done using the cosine rule

Put AB=1. We are looking for the locus of points like X which are k times as far from A as from B.

L and M are the points on the line AB which are on that locus.

LA:LB = k so if B is distance 1 from A, then L is at k/(k+1)

MA:MB = k so M is at k/(k−1)

K is midpoint of L and M so is at k2/(k2−1)

KB = k2/(k2−1) − (k2−1)/(k2−1) = 1/(k2−1)

Let θ = angle ABX

Cos rule in triangles ABX and KBX gives

k2x2 = 1 + x2 − 2x cos θ

r2 = x2 + 1/(k2−1)2 + 2x cos θ /(k2−1)

so

(k2−1) r2 = (k2−1)x2 + 1/(k2−1) + 2x cos θ

Adding the equations

(k2−1) r2 = 1 + 1/(k2−1) = k2/(k2−1)

so r = k/(k2−1)

and is the same for all X. So X follows a circle with centre B and radius k/(k2−1)  ▇