Complex plane line and ellipse distance Calculate the distance between the centre ellipse: 
$$2x^2+5y^2+20x-30y+15=0$$
and the line that is defined by the 2 centers of the following circles in complex plane.
$$ \left\lvert\ iz-3-4i\right\rvert=a $$
$$ \left\lvert\ iz-3z+20i \right\rvert=b$$    
I solved the centre of ellipse and got to: $2(x+5)^2+5(y-3)^2=80$ 
Which means that centre(-5,3). 
But how to get the centre of the complex equations and the distance to the centre of ellipse.
 A: Let $z=x+iy$ in $$\left\lvert\ iz-3-4i\right\rvert=a$$ and use the definition of norm of complex numbers, to find an equation in terms of $x$ and $y$. Similarly for the second circle find the equation and find the center.    
A: The first circle is the cirle defined by$$\left|z-\frac{3+4i}i\right|=a$$and the second one is the circle defined by$$\left|z+\frac{20i}{-3+i}\right|=\frac b{\sqrt{10}}.$$Can you take it from here?
A: To find the two centers, 
you need to use two key points.
1) In the complex plane, 
a circle is defined as
\begin{align}
\lvert z-c \rvert&=r
.
\end{align}
2) For $a,b\in\mathbb{C}$,
\begin{align}
|ab|&=|a|\cdot|b|
.
\end{align}
So, 
\begin{align}
a
&=|iz-3-4i|=|-i|\cdot|iz-3-4i|
\\
&=|(-i)(iz-(3+4i))|
\\
&=|z-(4-3i)|
,
\end{align}
so the first center is located at $(4,-3)$.
Similarly, for the other circle,
\begin{align} 
b&=
|iz-3z+20i|
=
|(i-3)z+20i|
\\
&=
\left\lvert(i-3)z+20i\cdot\frac{i-3}{i-3}\right\rvert
\\
&=
|i-3|\cdot
\left\lvert z-\frac{20i}{3-i}\right\rvert
,\\
\frac{b}{|i-3|}
&=
\left\lvert z-\frac{20i(3+i)}{(3-i)(3+i)}\right\rvert
\\
&=
\left\lvert z-\frac{20(-1+3i)}{10}\right\rvert
\\
&=
|z-(-2+6i)|
.
\end{align}  
Since we only need to find the center, we can leave 
left-hand side as it is,
and the second center is located at $(-2,6)$.
Given this three points, 
you have a completely defined triangle,
in which you just need to find one height.
