數痴夢王

Given two fixed points A and B. P is a variable point such that PAB=k:1.
Prove that the locus of P is a circle geometrially.

數痴夢王

not the stuff i want
Thanks anyway

數痴夢王

Solved but need cosine law

erectus

See attached.

數痴夢王

Do not think they are similar
as <POA = ,<POB
and <PAO less than <PBO

erectus

I don't see why you made such a simple mistake.

數痴夢王

Sorry Proved
2 sides proportional and included angle
so PA/PB = r/(r-1) = k

數痴夢王

erectus
不過這個証明假設了個軌跡是圓形，不能証明沒有其他圓形符合條件。

數痴夢王

放棄用幾何的方法証肥了，改用了解析幾何做
設 P(x,y)  A(0,0) B(a,0)
PA=k PB
x^2+y^2=k^2[(x-a)^2+y^2]
(k^2-1)x^2+(k^2-1)y^2-2k^2ax+k^2a^2=0
x^2+y^2-2k^2a/(k^2-1)x+k^2a^2/(k^2-1)=0
Proved r^2 = k^4a^2/(k^2-1)^2-k^2a^2/(k^2-1) = k^2a^2/(k^2-1)^2 >0
so it is a real circle.

erectus

No, it's already a proof. But with some details to be filled in.
In my diagram, it is assumed that k > 1.  (if 0 < k < 1, the circle will be around point A instead).
And it has also to be proved  when P is on the far end of the diameter, though it's quite easy.
This circle satisfies the locus of  point P. Can some point F lies inside or outside the circle still satisfies the condition?

Now AP:BP = k : 1.  Can FA be k(d-x) so that AF:BF= k : 1.? No, that should be FG which is parallel to AP.
So F cannot lie inside the circle.  It can be proved impossible outside the circle similarly. So the circle is the only locus.

數痴夢王

WHY FG is parallel to PA?

erectus

I mean FG parallel to AP in the diagram equals k(d-x) in length and cannot be equal to FA.

數痴夢王

got it thanks

erectus

you're welcome.

Here is another more definite proof: