# Need some help with Circle Equation of line to do with point P

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• May 29th 2011, 06:49 PM
insidesin
Need some help with Circle Equation of line to do with point P
Need help with question:
Quote:

Show that, whatever the position of P on the circle C, the point P’ representing z’ lies on a certain line, and determine the equation of this line:
I'm fairly clueless, other than the fact that I understand the line is probably linear below the x axis, because it's P'. (1/P)

Any help or ideas to get me moving further? P.s the questions before this one are as follows, with my answers as well:

Code:

```Item 1:     a)                  Sketch the circle C defined by the Cartesian co-ordinates:        (x, y):              x2 + (y – 1)2    =    1                   [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image001.png[/IMG]          From our knowledge of circles, we can express the co-ordinate equation in terms of the equation of a circle. (Ref:              Equation of a Circle)     Circle C co-ordinates:                        x2 + (y - 1)2 = 1     Equation of a Circle:              (x - h)2 + (y - k)2 = r2     (x – 0)2 + (y – 1)2  =  12     It can be seen that the following variables can be found simply by recognising their positions in comparison to the original circle equation.     h = 0    k = 1    and      r = 1     Where h and k are the x and y coordinates of the centre of the circle and r is the radius. We can then use the radius, x co-ordinate and y co-ordinate to sketch up the circle ‘C’:     [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image003.jpg[/IMG]                                 b)  The Point P(x,y) representing the non-zero complex number ‘z = x + iy’, lies on C. Express |z| in terms of Ѳ, the argument of z.     Sketch the diagram:                                       z  =      x  +  iy                               [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image004.png[/IMG]                            Ѳ                         [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image005.png[/IMG]                            z                                                       P                         [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image006.jpg[/IMG]     We need to find the Mod-arg form of |z| which is just the mod, so we’re finding the modulus through argument of z with use of imaginary number vector.     Since we know:    [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image008.png[/IMG]    can be written as:            [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image010.png[/IMG]     Where the complex number is expressed in polar (mod-arg) form.   We can then break the number up into its x and y components:     x component:   [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image012.png[/IMG]      y component:              [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image014.png[/IMG]     Now we can then go back and substitute each axis value into the original equation of the circle, this should represent the complex number as a form of the equation:                 Circle equation:   [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image016.png[/IMG]      [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image018.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image020.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image022.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image024.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image026.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image028.png[/IMG]     |z| expressed in terms of Ѳ is:      [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image030.png[/IMG]     Sub      [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image032.png[/IMG]    into      [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image034.png[/IMG] POSS REMOVE!         c)      Given that z’ =  [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image036.png[/IMG], find the modulus and argument of z’ in terms of Ѳ.     Since we know:   [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image038.png[/IMG]          and        [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image040.png[/IMG]     We can then substitute one into the other:     [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image042.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image044.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image046.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image048.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image050.png[/IMG]   Now that we have it in Mod-arg form, we can now express its argument and modulus singularly. But before we do this, it is important to put it into common form: ‘rCis[IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image052.png[/IMG]’ to make it clearer. We can do this with our knowledge of the unit circle, where the value of Cos(-Ѳ) = -Cos(Ѳ) and Sin(-Ѳ) = -Sin(Ѳ)     [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image050.png[/IMG]   [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image054.png[/IMG]   [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image056.png[/IMG]     We can now see that we have expressed z’ in Mod-arg form and can easily identify both the argument and th Since we can also state that the modulus of z’ isand the argument of z’ is [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image064.png[/IMG].     The diagram above just explains the inverse (or reciprocal) of z (which is z’) and where about they are on the plane with the conjugate of z as an extra vector to give extra reference inside the diagram.```

IN FACT, that didn't turn out so well, you can ignore it if you wish... or decipher it.

Questions:
Quote:

(iv) Sketch the curve C with Cartesian equation x2 + (y − 1)2 = 1. The point P,
representing the nonzero complex number z, lies on C.
(a) Express |z| in terms of θ, the argument of z.
(b) Given that z = 1
z find the modulus and argument of z in terms of θ.
Show that, whatever the position of P on the circle C, the point P representing z
lies on a certain line and determine the equation of this line.
Thanks!

p.s this editing form is really scrambled on my computer and it's hard to make this a neat post sorry in advance.