Given, $z_{1}=2+3 i$ and $z_{2}=3+4 i$
Now, we have $\left|z-z_{1}\right|^{2}+\left|z-z_{2}\right|^{2}=\left|z_{1}-z_{2}\right|^{2}$
(let $z=x+i y)$
$\Rightarrow|(x+i y)-(2+3 i)|^{2}+|(x+i y)-(3+4 i)|^{2}$
$\quad=|(2+3 i)-(3+4 i)|^{2}$
$\Rightarrow|(x-2)+i(y-3)|^{2}+|(x-3)+i(y-4)|^{2}$
$\quad=|-1-i|^{2}$
$\Rightarrow(x-2)^{2}+(y-3)^{2}+(x-3)^{2}+(y-4)^{2}=1+1$
$\Rightarrow \quad x^{2}+4-4 x+y^{2}+9-6 y$
$\quad+x^{2}+9-6 x+y^{2}+16-8 y=2$
$\Rightarrow \quad 2 x^{2}+2 y^{2}-10 x-14 y+36=0$
$\Rightarrow \quad x^{2}+y^{2}-5 x-7 y+18=0$
which represent a circle with centre $\left(\frac{5}{2}, \frac{7}{2}\right)$ and radius $\sqrt{\frac{25}{4}+\frac{49}{4}-18}=\frac{1}{\sqrt{2}}$