If $\arg\frac{z_1-z}{z-z_2}=\frac{\pi}{2}$, where $z_1=6+i$ and $z_2=4-3i$, then $\lvert z -(5-i)\rvert = \sqrt{5}$

Question

Let $z_1 = 6 + i$ and $z_2 = 4 - 3i$ . Let $z\in\mathbb{C}$ be a complex number such that $$\arg\left[\dfrac{z - z_1}{z_2 - z}\right] = \dfrac{\pi}{2}$$ Then, prove that $z$ satisfies $$\lvert z -(5-i)\rvert = \sqrt{5}$$

I don't know where I should begin.

Answer 1

The answer suggests that the solutions are on (a part of) a circle in the complex plane with Center $(5,-1)$ and radius $\sqrt{5}$. So let's write $z=a+bi$, here is an outline: Consider the complex number $\frac{6+i-a-bi}{a+bi-4+3i}$. If we multiply top and bottom by conjugate $(a-4)-(b+3)i$ and work out the numerator (that's just annoying algebra, nothing more), the numerator becomes of the form: $10a-2b-a^2-b^2-21+(4a-2b-22)i$. Now this complex number is supposed to end up on the positive "y-axis" as per given the argument. This implies that the real part is zero and the imaginary part $4a-2b-22>0$. The real part can be converted into a standard equation of a circle through Completing the Square (twice) and that equation is easily transformed into the equation in your question. Can you work out the details?

Answer 2

$\arg((z - z_1)/(z - z_2)) = \pm \pi/2$ means that $z - z_1$ is orthogonal to $z - z_2$ (their arguments differ by $\pm \pi/2 + 2 \pi k$). Therefore, $z$ lies on the circle for which $[z_1, z_2]$ is a diameter. The center of the circle is $(z_1 + z_2)/2$ and the radius is $|z_1 - z_2|/2$.