# Finding centre and radius of circle

Let $a,c \in \mathbb R$ with $a \neq 0$, and let $b \in \mathbb C$. Define $$S=\{z\in \mathbb C: az\bar{z}+b\bar{z}+\bar{b}z+c=0\}.$$

a. Show that $S$ is a circle, if $|b|^2 > ac$. Determine its centre and radius.
b. What is $S$ if $a=0$ and $b \neq 0$?

How would I get started in this? I'm completely stuck. Most appreciated.

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2
Start by writing $z=x+yi$, $b=p+qi$ and expand. – Henning Makholm Oct 20 '11 at 2:29

## 2 Answers

Where $z = x + iy$, your equation can be rewritten as $$a(x^2 + y^2) + 2Re(b\bar{z}) + c = 0$$ Writing $b = b_1 + b_2i$ this is the same as $$a(x^2 + y^2) + 2b_1x + 2b_2y + c = 0$$ Complete the square now....

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Thanks got it, appreciated! – John Southall Oct 21 '11 at 9:02

One way is to work your way backward to the problem from what you know about circles. A circle has a center and a radius. Let's call the center $w$, the radius, $r$. The circle is all the points $z$ whose distance from $w$ is $r$. The square of the distance between two points $u$ and $v$ in the complex plane is $|u-v|^2$. If $s$ is a complex number, then $|s|^2=s\overline s$. If you put all that together, you should come up with a formula looking very similar to the one in the problem, mostly just with different letters.

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