Complex locus help (1 Viewer)

[irvine]

A question from the terry lee book:



I get the shape of the locus, but am having trouble finding it's features (radius, centre ect)

Checked the "worked solution" but it didn't help/

Thanks

 

[Trebla]

From the x-intercepts of the curve, draw two lines which meet to where the centre of the circle should roughly be. Use some right angle trigonometry to deduce the length of the hypotenuse which is the radius (the line joining the centre and the x-intercept) and the y-value of the centre which is the side joining the centre and the x-axis (the x-value of the centre should be the midpoint of the x-intercepts if you remember from circle geometry).

 

[study-freak]

I got x^2+(y-1)^2=2, x^2+(y+1)^2=2 excluding (1,0), (-1,0).

 

[khorne]

Post working, study freak, it might help the OP more

 

[study-freak]

I didn't do it geometrically though.
I was waiting for someone (actually, OP) to confirm that it's right first.

 

[khorne]

Ok, my working:

arg(z-1) = 3pi/4 + arg(z+1) (p must lie above x axis)
or
arg(z-1) = -3pi/4 + arg(z+1) (p must lie below x axis)

therefore:

it can be shown that z+1 and z-1 meet at p at an angle of 3pi/4, thus p lies on a circle. From this, using the centre of the circle, and given the chord is length 2, you can find the y co-ord with trig.

But as the first one cannot be below x, y must be greater than 0.

do the inverse for the other one.

So my answers are the same as study freaks except the range is different:
x^2+(y-1)^2=2, x^2+(y+1)^2=2 excluding y <= 0 and y >=0

 

[Lukybear]

It can be done graphically, where two circles are drawn. The angle subtending at the circle = 1/2 the angle subtending at the centre. Hence, trignometric solving of radius + centre can be done.

 

[study-freak]

Ok, my working:

arg(z-1) = 3pi/4 + arg(z+1) (p must lie above x axis)
or
arg(z-1) = -3pi/4 + arg(z+1) (p must lie below x axis)

therefore:

it can be shown that z+1 and z-1 meet at p at an angle of 3pi/4, thus p lies on a circle. From this, using the centre of the circle, and given the chord is length 2, you can find the y co-ord with trig.

But as the first one cannot be below x, y must be greater than 0.

do the inverse for the other one.

So my answers are the same as study freaks except the range is different:
x^2+(y-1)^2=2, x^2+(y+1)^2=2 excluding y <= 0 and y >=0

But I have tested points on circles that are outside of your range, but are in my range and it worked.
Hence I think mine is right?

 

[khorne]

But I have tested points on circles that are outside of your range, but are in my range and it worked.
Hence I think mine is right?

If you consider it in two cases, and assume z (p) is below the x- axis then the z + 1 line must be rotated 3pi/4 to be parallel with z - 1. However, this would correspond to arg(z-1) = arg(z+1 ) - 3pi/4, but we needed at + 3pi/4 (for this case), thus, values below the x-axis do not work.

 

[study-freak]

If you consider it in two cases, and assume z (p) is below the x- axis then the z + 1 line must be rotated 3pi/4 to be parallel with z - 1. However, this would correspond to arg(z-1) = arg(z+1 ) - 3pi/4, but we needed at + 3pi/4 (for this case), thus, values below the x-axis do not work.

I get what you are saying and I knew that you were talking about this when you stated your answer above, but if that corresponds to arg(z-1) = arg(z+1 ) - 3pi/4, it still satisfies the original equation, right?

 

[khorne]

hmmm I suppose so.

 

[untouchablecuz]

 

[study-freak]

Yea, that's exactly what I did.
Except that there are two restrictions you need to add onto it.

 

[khorne]

Which I guess you can inspect, as 3pi/4 = 135, so if p is on 1,0 -1, 0 both the arguments need to be 180 or 0, so it does not hold.

 

[study-freak]

or undefined (argument)