Consider the equation of circle \(x^{2}\ -\ 16x\ +\ y^{2}\ -\ 10y\ +\ 73 = 0\)

To change the equation into standard form of the equation \((x\ -\ h)^{2}\ +\ (y\ -\ k)^{2} = r^{2}\)

Grouping the terms

\((x^{2}\ -\ 16x)\ +\ (y^{2}\ -\ 10y)\ +\ 73 = 0\)

Add and subtract 64 into \((x^{2}\ -16x)\)

and Add and subtract 25 into \((y^{2}\ -10y)\)

\((x^{2}\ -16x\ +\ 64\ -\ 64)\ +\ (y^{2}\ -\ 10y\ +\ 25\ -\ 25)\ +\ 73 = 0\)

\((x^{2}\ -\ 16x\ +\ 64)\ +\ (y^{2}\ -10y\ +\ 25)\ -\ 25\ -\ 64\ +\ 73 = 0\)

\((x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2}\ -16 = 0\)

\((x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2} = 16\)

Further simplify it

\((x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2} = 4^{2}\)

Which is standard form of the circle.

Compare it by standard form of the equation \((x\ -\ h)^{2}\ +\ (y\ -\ k)^{2} = r^{2}\)

\(h = 8,\ k = 5\text{ and } r = 4\)

Therefore, centre \((h,\ k) = (8,\ 5)\)

And radius \(r = 4\)

List four point on thw circle \((x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2} = 4^{2}\) is given below

\((8,\ 1),\ (12,\ 5),\ (8,\ 9)\text{ and }(4,\ 5)\)