 # A circle has the equation x^{2}+y^{2}-6x+12y+9=0. What is the di luipieduq3 2021-11-18 Answered
A circle has the equation $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}-{6}{x}+{12}{y}+{9}={0}$$. What is the distance from the center of the circle to the line $$\displaystyle{y}=-{2}{x}+{10}?$$

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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.