Find the centre and radius of $\arg{\left(\frac{z-(5+7i)}{z-(7+9i)}\right)}=\frac{\pi}{4}$?

Your idea is okay. Just recalculate carefully and you should be able to get the correct answer.

You can get the result by interpreting its geometric meaning without conducting tedious calculations.

Let us denote $Z(z), A(5+7i)$, and $B(7+9i)$. Since $\frac{z-(5+7i)}{z-(7+9i)} = \frac{5+7i-z}{7+9i-z}$, the given assumption is equivalent to $\angle BZA = \frac{\pi}{4}$. So by the inscribed angle theorem, $Z$ must be on the circle which has $AB$ as its chord.

There are two possible circles. Recall that $\arg$ is an "oriented" angle to determine the correct one.

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)