If ${\mathrm{z}}_1=8+4\mathrm{i}, {\mathrm{z}}_2=6+4\mathrm{i}$ and z be a complex
number such that Arg $\left(\frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{z}-{\mathrm{z}}_2}\right)=\frac{\mathrm{\pi }}{4}$, then the
locus of z is

$$\begin{gathered} {\mathrm{z}}_1=8+4\mathrm{i}, {\mathrm{z}}_2=6+4\mathrm{i} \\ \mathrm{given} \mathrm{that} \mathrm{Arg}\left(\mathrm{z}-{\mathrm{z}}_1\right) - \mathrm{Arg}\left(\mathrm{z}-{\mathrm{z}}_2\right) =\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{Arg}\left(\mathrm{x}+\mathrm{iy}-8-4i\right) -\mathrm{Arg}\left(\mathrm{x}+\mathrm{iy}-6-4i\right) =\frac{\mathrm{\pi }}{4} \\ \therefore {\mathrm{Tan}}^{-1}\left(\frac{\mathrm{y}-4}{\mathrm{x}-8}\right)-{\mathrm{Tan}}^{-1}\left(\frac{\mathrm{y}-4}{\mathrm{x}-6}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{{\displaystyle \frac{\mathrm{y}-4}{\mathrm{x}-8}} - {\displaystyle \frac{\mathrm{y}-4}{\mathrm{x}-6}}}{1+\left(\frac{\mathrm{y}-4}{\mathrm{x}-8}\right)\left(\frac{\mathrm{y}-4}{\mathrm{x}-6}\right)}=1 \\ \Rightarrow \left(\mathrm{y}-4\right)\left(x-6\right)-\left(\mathrm{y}-4\right)\left(x-8\right)=\left(x-6\right)\left(x-8\right)+{\left(\mathrm{y}-4\right)}^2 \\ \Rightarrow \left(\mathrm{y}-4\right)\left(\mathrm{x}-6-\mathrm{x}+8\right)={\mathrm{x}}^2-14\mathrm{x}+48+{\left(\mathrm{y}-4\right)}^2 \\ \Rightarrow {\mathrm{x}}^2-14\mathrm{x}+48+{\mathrm{y}}^2-8\mathrm{y}+16=2\left(\mathrm{y}-4\right) \\ \Rightarrow {\mathrm{x}}^2-14\mathrm{x}+48+{\mathrm{y}}^2-10\mathrm{y}+24=0 \\ \Rightarrow \left({\mathrm{x}}^2-14\mathrm{x}+49\right)+\left({\mathrm{y}}^2-10\mathrm{y}+25\right)=2 \\ \therefore \boxed{{(\mathrm{x}-7)}^2+{(\mathrm{y}-5)}^2=2} \end{gathered}$$