# Question about circle chord midpt locus

1. Jan 10, 2013

### kurketom

1. The problem statement, all variables and given/known data

A is the pt where the circle with wquation x^2+y^2=25 cuts the positive x-axis. Find the midpts of the chords of this circle that contain the pt A

2. Relevant equations

3. The attempt at a solution

Since it is about the midpt of chords, I try to set up a equation for the chords:

y/x-5 = (√(25 - a^2)) / ((√(25 - b^2) - 5)

where (a,b) are the pt on the circle and their ranges are -5<=a<=5 -5<=b<=5 and (x,y) are the pt will fit in the chord

Then I used another equation which is perpendicular to the chord and pass through the center of the cirlce (0,0):

y/x = (5 - (√(25 - b^2)) / (√(25 - a^2))

Since the intersection of these two pts will be the mid pt of the chord by combining them together it should get the locus. But turn out to be very wired and wrong.... Please help me out!

2. Jan 10, 2013

### Dick

You can represent a point on the circle as (t,sqrt(25-t^2)). The point A is (5,0). You can find the midpoint of that chord without any line equations. Just take the sum and divide by 2.

Last edited: Jan 10, 2013
3. Jan 10, 2013

### Vargo

This problem is simpler if you abandon coordinates and recognize it as a homothetic transformation with center A and scale factor of 1/2. Do you know what similar figures are in Euclidean geometry? Even if you want to express your final answer in terms of an equation it will be simpler to visualize the answer and then write down the equation than it would be to derive the equation algebraically. (though I grant that your instructor might prefer the algebraic method).

http://en.wikipedia.org/wiki/Homothetic_transformation

4. Jan 10, 2013

### kurketom

Man can't believe the answer is so simple.... Anyway thx a lot!!!! And homothetic transformation never heard of it! Thx for letting me know! I will keep reading. But yeah I think I will just answer with the algebraic method first. Thx again for the replies!