The equation of the circle centered at $1+2i$ with radius $1$ can be written as \[ z\overline{z} + pz + q \overline{z} + r = 0.\]What is the ordered triple $(p, q, r)$?
+118608 Stop being so lazy and at least attempt to make it readable.
Put you more difficult LaTex into the LaTex box and see if you can make it read properly.
try deleting the bits that come up red.
+118608 Thanks Tertre but you missed my point. I could do that too. In fact I did do that and then I changed my mind.
If people do not put effort into presenting their questions properly (or at least as good as they are capable of) then perhaps some other people cannot be bothered answering.
Latex is computer code used for displaying mathematics.
The latex box is in the ribbon. It is the button that say LaTex. It is there when you are writing any post.
Click on it to open it.
then copy some of your code and paste it in. See what happens.
It is easy to learn a little bit of Latex. And when you know a little bit it is easy to learn more.
You will find it very helpful to become comfortable with very basic coding.
Your post is displaying better now, which is odd because I see no evidence that anyone has edited it. Perhaps it was better than I could see originally and just was not displaying properly.
You can still get rid of the meaningless $ signs though - That would make the question read better.
+4609 Here, I'll do it: The equation of the circle centered at \(1+2i\) with radius \(1\) can be written as \(z\overline{z} + pz + q \overline{z} + r = 0.\)What is the ordered triple \((p,q,r)\)?
+118608 Maybe this will help. 
I have not watched it but I would have to learn how to do this befor I could teach you and it is Internet sites, often youtube clips, that I learn from.
I just googled "How do I graph a circle in the complex plane." And this was the first clip that appeared.
+26367 circle:
\(|z-z_0|=1\\ if~z=x+yi ~and~z_0=1+2i\\ then~(x-1)^2+(y-2)^2=1\\ or~x^2+y^2-2x-4y+4=0\)
Formula:
\(z\overline{z}+pz+q\overline{z}+r=0\\ if~z=x+yi~and~\overline{z}=x-yi\\~and~z_0=1+2i\\ then~x^2+y^2+(p+q)x+(pi-qi)y+r=0\)
p,q = ?
\(p+q=-2\\\ (p-q)i=-4\\ r=4\\ \Rightarrow \\ q=-1-2i\qquad p=-1+2i\qquad r=4\)
(p,q,r) = ( -1+2i, -1-2i, 4 )
