Find the centre and radius of $\arg{\left(\frac{z-(5+7i)}{z-(7+9i)}\right)}=\frac{\pi}{4}$? I'm having huge trouble trying to find the centre and radius of the circle that contains the segment $\arg{\left(\frac{z-(5+7i)}{z-(7+9i)}\right)}=\frac{\pi}{4}$
At first I tried to solve for the intersection of the two lines running from the centre to $(5,7)$ and $(7,9)$. That is, I tried to solve for the intersection of $(y-9)=m_1(x-7)$ and $(y-7)=m_2(x-5)$, but there are too many unknown variables so that didn't work.
Next I tried to find the cartesian equation to $\arg{\left(\frac{z-(5+7i)}{z-(7+9i)}\right)}=\frac{\pi}{4}$ by letting $z=x+iy$.
After substituting that in and realising the denominator I got $\arg{\left(\frac{(x-5)(x-7)-(y-7)(y-9)}{(x-7)^2+(y-9)^2}+i\frac{(x-5)(y-9)+(y-7)(x-7)}{(x-7)^2+(y-9)^2}\right)}=\frac{\pi}{4}$
Which I think then means $\tan^{-1}{\left(\frac{(x-5)(y-9)+(y-7)(x-7)}{(x-5)(x-7)-(y-7)(y-9)}\right)}=\frac{\pi}{4}$
Which gave the curve $(x-5)(y-9)+(y-7)(x-7)=(x-5)(x-7)-(y-7)(y-9)$, however when I plugged that into desmos I got a hyperbola with the minimum of the upper branch being $(7,9)$ and the max of the lower branch being $(5,7)$. So I've clearly done something wrong when finding the cartesian equation.
Nevertheless does anyone know the most efficient way to find the centre and radius? I'm not well versed in circle geometry theorems so maybe there is something there I have missed?

Your mistake is in simplifying the contents of the arg bracket.
The real part is $$\frac{(x-5)(x-7)+(y-7)(y-9)}{(x-7)^2+(y-9)^2}$$ and the imaginary part is $$\frac{(y-7)(x-7)\color{red}{-}(x-5)(y-9)}{(x-7)^2+(y-9)^2}$$
Since the arg is $\frac{\pi}{4}$ we can set these equal to each other and this leads to the equation of the circle $$(x-7)^2+(y-7)^2=4,$$
with the obvious conclusion.
The result can be arrived at very easily just by drawing a picture and no algebra (as has been indicated by others)
