locus

[Jeremy]

Hi everyone,
I have been trying to do this question and can't get the answer in the back of the book. Please help. I have spent hours doing it.
- Jeremy

A is a point where the circle with equation x^2 +y^2 = 16 cuts the x-axis. Find the locus of the midpoints of all chords of this circle that contain the point A.

I have tried to allow a point P(x,y) as the midpoint of the chord and found PA^2 to be (x-4)^2 + (y-0)^2 but don't know what to do from here.

It is probably easy but I just can't see it.

The answer is meant to be x^2 + y^2 - 4x =0

 

[tkhunny]

The problem statement doesn't say how it decided which point to pick. There are two. However, let's go with (4,0).

The midpoints are averages of coordinates of any point on the circle and the point (4,0). This gives an expression for the desired locus,

[(x+4)/2 , y/2]

Just worying about the positive half (the piece above the x-axis, this can be expressed in terms of 'x' only.

[(x+4)/2 , sqrt(16 - x^2)/2]

This is a little sneaky. We now have f((x+4)/2) = sqrt(16 - x^2)/2
A change of variables, z = (x+4)/2 gives x = 2*z - 4.
Substituting this into the definition of 'f', gives

f(((2*z - 4)+4)/2) = sqrt(16 - (2*z - 4)^2)/2

Simplifying:

f(z) = sqrt((4-z)*z)

If we then plot the 'z' variable on the same axis as the previous 'x' variable, we get y = sqrt((4-x)*x), or, since we wish to consider both above and below the x(z)-axis y^2 = 4*x - x^2.

 

[soroban]

Hello, Jeremy!

I already knew the answer to this problem,
. . but I too had a difficult time getting to the equation.

A is a point where the circle with equation x² + y² = 16 cuts the x-axis.

Find the locus of the midpoints of all chords of this circle that contain the point A.

Let point A be (4,0).

Let B be any poi<u>nt on </u>the circle: .(P,Q),
. . where .Q = √16 - P²

Then midpoint M of segment AB is: .([P+4]/2, Q/2)

The locus is:. x .= .[P+4]/2 .and .y .= .Q/2


We have: .[1] P .= .2x - 4 .and .[2] 2y .= .√(16 - P²)

Substitute [1] into [2]: . 2y .= .√[16 - (2x - 4)²]

. . which simplifies to: .x² - 4x + y² .= .0

This is the circle: .(x - 2)² + y² .= .4 . . . with center (2,0) and radius 2.