The equation of the diameter of the circle ${(x-2)}^2+{(y+1)}^2=16$ which bisects the chord cut of by the line $x-2y-3=0$

Given Equation is ${(\mathrm{x}-2)}^2+{(\mathrm{y}+1)}^2=16$

Equation of chord is $\mathrm{x}-2\mathrm{y}-3=0$

slope of chord=$\frac{1}{2}$

Since  diameter bi-sects the chord then 

Requqired line is passing through (2,-1) and perpendicular to given chord

Then the Equation of  the line passing through (2, -1) and slope $=-2$is

$$\begin{gathered} \mathrm{y}+1=-2(\mathrm{x}-2) \\ \mathrm{y}+1=-2\mathrm{x}+4 \\ 2\mathrm{x}+\mathrm{y}-3=0 \end{gathered}$$