Explanation:
D Let the other end (n, k)
\(x^2+y^2-8 x-4 y+c=0\)
general from of circle
\(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\)
center \((-\mathrm{g},-\mathrm{f})\)
Comparing equation (i) and (ii) we have
\(\therefore 2 \mathrm{gx}=-8 \mathrm{x}, \quad 2 \mathrm{fy}=-4 \mathrm{y}\)
\(\begin{array}{cl}2 \mathrm{~g}=-8 2 \mathrm{f}=-4 \\ \mathrm{~g}=-4 \mathrm{f}=-2 \\ \text { center, }(-\mathrm{g},-\mathrm{f}) \equiv(4,2)\end{array}\)

Mid-point form at \(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\)
\((4,2)=\frac{-3+\mathrm{n}}{2} , \frac{2+\mathrm{k}}{2}\)
\(\frac{-3+\mathrm{n}}{2}=4 , \frac{2+\mathrm{k}}{2}=2\)
\(-3+\mathrm{n}=8 , 2+\mathrm{k}=4\)
\(\mathrm{n}=8+3 , \mathrm{k}=4-2\)
\(\mathrm{n}=11 , \mathrm{k}=2\)So, \(\quad(\mathrm{n}, \mathrm{k}) \equiv(11,2)\)