Where is the Circle

A point Q moves in such a way that its distance from the point T(2,-3) is twice its distance from the point S(0,3). Find the equation of the locus of point Q.

This is a typical Form 4 Coordinate Geometry question. If you understand the concepts correctly, you should be able to get 3x^2 + 3y^2 + 4x - 30y + 23 = 0.

I still remember the form 4 textbook asks "can you tell the shape of the locus from the equation?" Honestly, I didn't know what the shape is. I reckoned it should be some sort of peculiar curve, though later I was kind of convinced that it should be a circle, though I did not attempt to prove it.

I came across this topic again in Form 6 Mathematics. According to the Maths teacher, we can know that it is a circle because

(i) the coefficients of x^2 and y^2 are equal.
(ii) there is no xy term.

Maybe I was not paying attention, but another condition is that the (radius)^2 cannot be negative.

After generalising the above question to:

A point Q moves in such a way that its distance from the point A(a,b) is k times its distance from the point S(c,d). Find the equation of the locus of point Q.

Hint : 1. The method is the same, just that we use algebraic expressions in place of numerical figures.
2. Use completing the square. Complete the squares of x^2 and y^2.
3. Move the (x+something)^2 + (y+another thing)^2 term to LHS.
4. To prove that the locus is a circle, we need to show that RHS is positive. This can be done by a bit of algebraic manipulations.

The end result:
The equation of locus
(x-a)^2 + (y-b)^2 = (k^2) [(x-c)^2 + (y-d)^2] can also be expressed as

[x + (a-ck^2)/(k^2 - 1)]^2 + [y + (b-dk^2)/(k^2 - 1)]^2 =
{(k^2)[(a-c)^2 + (b-d)^2]}/(k^2 - 1)^2 , k not = +/- 1.

As you can probably see, the equation of locus is also an equation of a circle.

I have also shown (not here) that the centre of the circle is on the line joining A and B.


References:
1. Maths lecture in school, 福福, 11/10/2009.
2. Pre-U Text STPM Mathematics S&T Paper 1, Tai Pon Hoy, Longman, chapter 5.
3. Success Series, Additional Mahtematics SPM, Wong and Wong, Oxford Fajar, pg.123.

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Anyway, I still remember in form 3, prior to PMR, when I was still struggling with my sexual orientation, the KH teacher asked the whole class, among triangle, square, circle, which is our favourite and gave us 3 seconds to think.

I cannot decide between triangle and circle. I don't like circle that time, because I don't like pi. Even until today I still don't know how to get the value of pi and how to prove the area of circle is (pi)r^2. But I also don't like trigonometry, and I recalled that some Greek mathematician or philosopher said that circle is the most perfect shape, so in the end I chose circle.

The teacher said that all who chose circle are gila seks. I was quite happy; since I'm gila seks I cannot be gay, I thought.

It was later that I came to realize men also have circles.

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* If a rectangle is analogous to an area covered by the linear movement of a line segment in the direction perpendicular to itself, what is analogous to a sphere? A volume covered by the rotational movement of a circle about an axis which is on the same plane as the circle and passing through the circle's centre?

闭门造车。学而不思则x, 思而不学则y.

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