If $z_1=10+6i, z_2=4+6i$ and z is complex number such that $\left|z-7-9i\right|=k$ and $amp\left(\frac{{\displaystyle z-z_1}}{{\displaystyle z-z_2}}\right)=\frac{\pi }{4}$ then $k^2=\mathrm{.......}$

Given $z_1=10+6i, z_2=4+6i$ and let $z=x+iy$

$\frac{z-z_1}{z-z_2}=\frac{(x+iy)-(10+6i)}{(x+iy)-(4+6i)}=\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)}=\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)}\times \frac{(x-4)-i(y-6)}{(x-4)-i(y-6)}$

$=\frac{[(x-10)(x-4)+{(y-6)}^2]+i[(x-4)(y-6)-(x-10)(y-6)]}{{(x-4)}^2+{(y-6)}^2}$

$=\frac{(x^2+y^2-14x-12y+76)+i(6y-36)}{{(x-4)}^2+{(y-6)}^2}$

$amp\left(\frac{{\displaystyle z-z_1}}{{\displaystyle z-z_2}}\right)=\frac{\pi }{4}\Rightarrow \frac{6y-36}{x^2+y^2-14x-12y+76}=1\Rightarrow x^2+y^2-14x-12y+76=6y-36$

$\Rightarrow x^2+y^2-14x-18y+112=0\Rightarrow {(x-7)}^2+{(y-9)}^2=18$

$k^2={|z-7-9i|}^2={|(x-7)+i(y-9)|}^2={(x-7)}^2+{(y-9)}^2=18$