Complex plane line and ellipse distance

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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.