1. ## Cartesian equation

I need help to find the cartesian equation for the region represented by

| z−2+i | = 1/5 | z−8−3 i |

The question asks me to do the following:Please put your answer in a "natural" form. If you haven't already done so, complete the square for the terms in x and complete the square for the terms in y, and hence deduce that the locus of points satisfied by the relation is associated with a circle. What is the centre of that circle?
Any help thanks in advance .

2. Hello, skirk34!

Find the cartesian equation for: .$\displaystyle | z-2+i | \:= \:\tfrac{1}{5}| z-8-3 i |$

We have: .$\displaystyle 5|z - (2-i)| \:=\:|z - (8+3i)|$

We want the points $\displaystyle P(x,y)$ so that the distance from $\displaystyle P$ to (8,3)
. . is five times the distance from $\displaystyle P$ to (2,-1).

We have: .$\displaystyle 5\sqrt{(x-2)^2 + (y+1)^2} \;=\;\sqrt{(x-8)^2 + (y-3)^2}$

Square both sides: .$\displaystyle 25\left[(x-2)^2 + (y+1)^2\right] \;=\;(x-8)^2 + (y-3)^2$

. . which simplifies to: .$\displaystyle 24x^2 - 84x + 24y^2 + 56y \;=\;-52$

Divide by 24: . $\displaystyle x^2 - \frac{7}{2}x + y^2 + \frac{7}{3}y \;=\;-\frac{13}{6}$

Complete the square: . $\displaystyle x^2 - \frac{7}{2}x + {\color{blue}\frac{49}{36}} + y^2 + \frac{7}{3}y + {\color{red}\frac{49}{36}} \;=\;-\frac{13}{6} + {\color{blue}\frac{49}{16}} + {\color{red}\frac{49}{36}}$

. . And we have: . $\displaystyle \left(x - \frac{7}{4}\right)^2 + \left(y + \frac{7}{6}\right)^2 \;=\;\frac{325}{144}$

We have a circle . . . .center $\displaystyle \left(\frac{7}{4},\:-\frac{7}{6}\right)$ and radius $\displaystyle \frac{5\sqrt{13}}{12}$

But check my algebra and arithmetic . . . please!
.

3. Your algebra is good, thanks