Think More About the HK Reunion

James Koo

Peter Tong's article about Appollonius Circle aroused my interest because it appeared on the same day of the article that I tried to keep the interest of HK reunion high. I know all Pui Ching graduates are very strong in math, and they all will spend the time trying to solve these problems instead of looking at the reunion article. Therefore I decided to give a try to solve one, so other Lighters can spend more time to read and think about the re-union. Another reason is that recently a Lighter called me a businessman, the lowest rank (in China) among scholars, bureaucrats, laborer and businessman, and I will give it a try again to join the rank of scholar. So here it is.

The Problem:

Prove: The locus for all points with the distances to points A and B with ratio k while k<>1 is a circle.

Proof:

I will give a conceptual proof first for those who do not wish to read the math expressions. Basically first find the expression for the locus, and then find this is a circle. 

I will use +, -, *, and /, for addition, subtraction, multiplication and division as there is no symbol for division on the keyboard. I will use x*x for x square and state square root.

First put the points A and B with A at the origin, and B on the Y-axis with a distance "a" from the origin.

The points x=a*k/(k+1) and x=a*k/(k-1) on the Y-axis will meet the ratio condition.

If these two points are a part of circle, then the center will be at x=a*k*k/(k*k-1) on the Y-axis and the radius is a*k/(k*k-1).

The distance of any point x, y to point A is square root of x*x + y*y, and to point B is the square root of (x-a)*(x-a) + y*y. 

Therefore the expression for the loci is:

Square root of (x*x + y*y) / square root of ((x-a)*(x-a) + y*y)) = k

Or 

(x*x + y*y)/ ((x-a)*(x-a) + y*y) = k*k as we know both sides are positive.

Or

y*y = (x*x-k*k(x-a)*(x-a))/(k*k-1)

Now the equation for the circle with center at x=a*k*k/(k*k-1) and radius = a*k/(k*k-1) is:

(x-a*k*k/(k*k-1))*(x-a*k*k/(k*k-1)) + y*y = (a*k/(k*k-1))*(a*k/(k*k-1))

You will find the expression of y*y of the locus will satisfy the above equation.

QED.

If you need more detail, I will send to you upon request.

Actually when k=1 is a special case as a straight line is a circle with infinite radius.

Now please go back read the reunion articles and come to HK.