# Write the equation of the circle with center at (-4,\ -5) and to

Write the equation of the circle with center at $$\displaystyle{\left(-{4},\ -{5}\right)}$$ and touching the line $$\displaystyle{3}{x}-{2}{y}-{7}={0}$$

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Ched1950

Step 1
Consider that the center of circle be $$\displaystyle{\left(-{4},\ -{5}\right)}$$ and line $$\displaystyle{3}{x}-{2}{y}-{7}={0}$$ touches the circle.
Since the line $$\displaystyle{3}{x}-{2}{y}-{7}={0}$$ touches the circle implies the shortest distance of the centre and line is the radius.
Shortest distance of a line $$\displaystyle{A}{x}+{B}{y}+{C}={0}$$ from the point $$\displaystyle{\left({a},\ {b}\right)}$$ is given by: $$\displaystyle{\left|{\frac{{{A}{a}+{B}{b}+{C}}}{{\sqrt{{{A}^{{{2}}}+{B}^{{{2}}}}}}}}\right|}$$
Therefore,
$$\displaystyle{r}={\left|{\frac{{{3}{\left(-{4}\right)}-{2}{\left(-{5}\right)}-{7}}}{{\sqrt{{{3}^{{{2}}}+{\left(-{2}\right)}^{{{2}}}}}}}}\right|}$$
$$\displaystyle={\left|{\frac{{-{12}+{10}-{7}}}{{\sqrt{{{9}+{4}}}}}}\right|}$$
$$\displaystyle={\frac{{{9}}}{{\sqrt{{{13}}}}}}$$
Step 2
Equation of circle with center $$\displaystyle{\left({a},\ {b}\right)}$$ and radius r is given by $$\displaystyle{\left({x}-{a}\right)}^{{{2}}}+{\left({y}-{b}\right)}^{{{2}}}={r}^{{{2}}}$$
Since centre is $$\displaystyle{\left(-{4},\ -{5}\right)}$$ and radius is $$\displaystyle{\frac{{{9}}}{{\sqrt{{{13}}}}}}$$. Therefore, equation of circle become:
$$\displaystyle{\left({x}-{\left(-{4}\right)}\right)}^{{{2}}}+{\left({y}-{\left(-{5}\right)}\right)}^{{{2}}}={\left({\frac{{{9}}}{{\sqrt{{{13}}}}}}\right)}^{{{2}}}$$
$$\displaystyle{\left({x}+{4}\right)}^{{{2}}}+{\left({y}+{5}\right)}^{{{2}}}={\frac{{{81}}}{{{13}}}}$$