Cartesian equation

skirk34 · #1 $Old$ October 15th, 2008, 08:30 AM

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

.

Soroban · #2 $Old$ October 15th, 2008, 02:45 PM

Hello, skirk34!

Quote:

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

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

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

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

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

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

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

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

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

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

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

skirk34 · #3 $Old$ October 15th, 2008, 08:56 PM

Your algebra is good, thanks