Given circle is $x^{2 }+ y^{2} - 16 = 0$
Eqn. of chord say AB of given circle is
$3x + y + 5 = 0$.
Equation of required circle is
$x^{2} +y^{2}-16 + \lambda\left(3x+y + 5\right) =0$
$\Rightarrow x^{2} + y^{2} + \left(3\lambda\right)x+\left(\lambda\right)y+5\lambda -16 =0\quad...\left(1\right)$
Centre $C = \left(\frac{-3\lambda}{2},\frac{-\lambda }{2} \right)$
If line AB is the diameter of circle $\left(1\right)$, then
$C \left(\frac{-3\lambda }{2},\frac{-\lambda }{2} \right)$ will lie on line AB.
$3\left(\frac{-3\lambda }{2}\right)+\left(\frac{-\lambda }{2}\right) + 5 = 0$
$\Rightarrow -\frac{9\lambda-\lambda}{2}+5 = 0 \Rightarrow \lambda = 1$
Hence, required eqn of circle is
$x^{2} + y^{2} +3x +y+ 5-16 = 0$
$\Rightarrow x^{2} +y^{2} +3x + y-11 = 0$