Line and Circle Loci
By Niranjan Srinivas on Saturday, September 08, 2001 - 07:34 am:

1) Two different lines and a point 0 are fixed in the plane. A variable line is drawn through O to cut the two fixed lines at points R and S. Define a point P on RS such that 2/OP = 1/OR + 1/OS. Show that the locus of P is a line.

2) Find the equation of the circle passing through the points of intersection of the circles given by equatioins x2 + y2 -8x-6y-3 = 0 and x2 + y2 +7x+9y-11 = 0 and through the point (-3,5).

For 1) I had a few ideas but nothing seemed to work out. I intitially chose O as the origin but later I neglected that and chose one of the fixed lines as the x axis, but that didn't work either. Please help!
For 2) The circle required is unique as it passes through three points. My idea was to solve the first two equations to get the two points and then it would be plain sailing, but how to solve two simultaneous quadratics in x and y ? Or is there a better method which does not require such solving ?
Thanks

Niranjan.

By Michael Doré on Saturday, September 08, 2001 - 05:51 pm:

For 2), there may well be a better method, but your method will work without too much trouble. To find the points of intersection we need to solve:

x2 + y2 - 8x - 6y - 3 = 0
x2 + y2 + 7x + 9y - 11 = 0

If you subtract these two equations, you can cancel the x2 + y2 and get a relationship of the form y = ax + b. You can then eliminate y in one of the original equations, and this leaves you with a quadratic in x to solve. There are two solutions for x, and for each one you can find the corresponding value of y by plugging x into the linear relation. Once you have the points of intersection, as you say, it's plain sailing.

By Kerwin Hui on Sunday, September 09, 2001 - 03:53 pm:
For 1, it may be easier if you know the polar equation, and the normal form of a line (assuming you are working in 2-D Euclidean space), but the solution is still a bit messy.

Here is the solution: I choose my origin to be the intersection of the two fixed lines (lines parallel is trivial) and my solution relies on the normal form

x cos a+ y sin a- p = 0

where a is the angle made by the normal of the line and the x-axis in the usual direction sense and p is the displacement of the foot of perpendicular from O (signs are important here), and after a long and messy calculation, I made the locus of P to be a straight line in polar coordinates.

Kerwin

By David Loeffler on Thursday, September 13, 2001 - 04:40 pm:
There is a synthetic proof. This is hopelessly technical but quick. It may be wrong, so feel free to criticise.

Inversion looks promising. If we invert in a fixed circle centre O, then the line loci of R and S invert into two circles both passing through O. Crucially, P ' is now the midpoint of R ' and S ' , and R ' S ' is still a straight line passing through O.

The canonical polar equation of a circle is r = A cos (q-a). So, taking O as the origin, we obtain r P ' = 1/2 (r R ' + r S ' ) = B cos (q- b) for some B and b. Thus the locus of P ' is also a circle through O, which will invert back into a line. QED.

David

By Arun Iyer on Thursday, September 13, 2001 - 08:13 pm:

David,
I was trying to ponder on what you were saying, but it is something that I have never come across. What is inverson?
love arun

By David Loeffler on Friday, September 14, 2001 - 10:37 am:

As I said, it's technical. There is a description of inversion here .

[Editor's Comment: Complex inversion has been discussed before in NRICH, see 1/z transformation . With this tool, David's proof is excellent.]
By Niranjan Srinivas on Friday, September 14, 2001 - 12:19 pm:

Hello ! No inversion or vectors necessary !
I asked my Maths Professor who gave me a simple solution. Here it goes:
Choose O as Origin.

Let the angle the variable line makes with the positive X axis be q. Then equation of the line is y = x tanq

P is a point on R S.

Also, 2/O P = 1/O R + 1/R S.

To find the locus of P.

Clearly R has coordinates

(O R cosq,O R sinq)

This lies on a fixed line whose equation is,say,

l x + m y = 1 (on dividing by the constant and taking THAT as the constant)

=> l.O R.cosq+ m.O R.sinq = 1

=> 1/O R = l cosq+ m sinq

||l y a cosq+ b sinq = 1/O S

where a, b are the constants in the equation of the other fixed line a x + b y = 1.

Adding,

(l+a)cosq+ (m+b)sinq =

1/O R + 1/O S = 2/O P.

Therefore

(l+a)O P cosq+ (m+b)O Psinq = 2

Locus of P is

(l+a)x + (m+b)y = 2

because O P cosq = x coordinate of P

and O P sinq = y coordinate of P

Locus of P is a line because (l+a)x + (m+b)y = 2 represents a straight line !
Terribly simple, isn't it ?

Cheers
Niranjan.

By David Loeffler on Friday, September 14, 2001 - 12:56 pm:

Yes, but it's also exactly equivalent to my solution above (mine just dresses the same algebra up in geometric language).

David

By Arun Iyer on Monday, September 17, 2001 - 07:34 pm:

One thing that comes out of the proof by Niranjan's maths professor is that the given question can be further generalised by stating....
P is a point such that
p/OR + q/OS = r/OP .........
where p,q and r are arbitrary non-zero constants.
Still the locus of point P is a line!

Could this be quite useful somewhere?
love arun