complex q. (1 Viewer)

[spikestar]

i have been trying to do this over an hour now and i cant figure it out. i'm supposed tio make it into graph it and make it into and catesian eq. can some 1 do it plz so i can c what i'm doing wrong

 

[Mountain.Dew]

i wont tell u the answer, BUT i will tell you how u should approach it...

think about it this way: |z-@| / |z-$| = |(z-@)/(z-$)| <== then u 'realise' the denominator once u substitute z=x+iy, then square both sides of equation, and u have ur locus. fiddle around with algebra a bit, and it should turn out nice.

 

[Riviet]

Most straight forward way of doing this is let z=x+iy, then use the fact that |a+ib|=sqrt(a²+b²), square both sides, then multiply the denominator across to the 2, expand everything, simplify a bit, then complete the square and you end up with the locus of a circle with centre (2,1) and radius 2 units.

 

[spikestar]

thank you