Step 1

The equation of the circle is given as,

\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}-{6}{x}+{12}{y}+{9}={0}\)

The general equation of the circle with center at \(\displaystyle{\left({h},\ {k}\right)}\) and radius r is given as,

\(\displaystyle{\left({x}-{h}\right)}^{{{2}}}+{\left({y}-{k}\right)}^{{{2}}}={r}^{{{2}}}\)

On simplifying the equation of the circle given as,

\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}-{6}{x}+{12}{y}+{9}={0}\)

\(\displaystyle{\left({x}^{{{2}}}-{6}{x}\right)}+{\left({y}^{{{2}}}+{12}{y}\right)}+{9}={0}\)

\(\displaystyle{\left({x}^{{{2}}}-{2}\times{x}\times{3}+{3}^{{{2}}}-{3}^{{{2}}}\right)}+{\left({y}^{{{2}}}+{2}\times{y}\times{6}+{6}^{{{2}}}-{6}^{{{2}}}\right)}+{9}={0}\)

\(\displaystyle{\left({x}^{{{2}}}-{2}\times{x}\times{3}+{3}^{{{2}}}\right)}+{\left({y}^{{{2}}}+{2}\times{y}\times{6}+{6}^{{{2}}}\right)}+{9}-{3}^{{{2}}}-{6}^{{{2}}}={0}\)

\(\displaystyle{\left({x}-{3}\right)}^{{{2}}}+{\left({y}+{6}\right)}^{{{2}}}-{36}={0}\)

\(\displaystyle{\left({x}-{3}\right)}^{{{2}}}+{\left({y}-{\left(-{6}\right)}\right)}^{{{2}}}={\left({6}\right)}^{{{2}}}\)

On comparing the equations, we get the center of the circle as \(\displaystyle{\left({3},\ -{6}\right)}\) and the radius of the circle as 6 units.

Step 2

The equation of the line from which the distance between center and the line is to be calculated is,

\(\displaystyle{y}=-{2}{x}+{10}\)

The general equation of the line is,

\(\displaystyle{y}=-{2}{x}+{10}\)

\(\displaystyle{2}{x}+{y}-{10}={0}\)

The distance of line \(\displaystyle{a}{x}+{b}{y}+{c}={0}\) from the point \(\displaystyle{\left({x}_{{{1}}},\ {y}_{{{1}}}\right)}\) is given as,

\(\displaystyle{d}={\frac{{{a}{x}_{{{1}}}+{b}{y}_{{{1}}}+{c}}}{{{\left|\sqrt{{{a}^{{{2}}}+{b}^{{{2}}}}}\right|}}}}\)

Putting the center and the equation of the line, we get

\(\displaystyle{d}={\left|{\frac{{{2}\times{\left({3}\right)}+{\left(-{6}\right)}-{10}}}{{\sqrt{{{\left({2}\right)}^{{{2}}}+{\left({1}\right)}^{{{2}}}}}}}}\right|}\)

\(\displaystyle={\left|{\frac{{-{10}}}{{\sqrt{{{5}}}}}}\right|}\)

\(\displaystyle={2}\sqrt{{{5}}}\)

Therefore, the required distance between the center of the circle and the given line is \(\displaystyle{2}\sqrt{{{5}}}\) units.