newbie qn (1 Viewer)
- Thread starter sven0023
- Start date
spice girl
magic mirror
- Joined
- Aug 10, 2002
- Messages
- 785
locus is
(x - 1/2)^2 + (y - 1)^2 = 5/4
(x - 1/2)^2 + (y - 1)^2 = 5/4
standard method -
let z = x + iy then:
(x+iy-2i)/(x-1+ iy) = imaginary
times top and bottom by (x-1-iy)
then Re[(x+iy-2i)(x-1-iy)] =0
ie. x^2 - x + y^2 - 2y = 0
complete square
(x-1/2)^2 + (y-1)^2 = 1+ 1/4 = 5/4
like spice girl's answer
let z = x + iy then:
(x+iy-2i)/(x-1+ iy) = imaginary
times top and bottom by (x-1-iy)
then Re[(x+iy-2i)(x-1-iy)] =0
ie. x^2 - x + y^2 - 2y = 0
complete square
(x-1/2)^2 + (y-1)^2 = 1+ 1/4 = 5/4
like spice girl's answer