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

enfurezca3x 2021-11-15 Answered
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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Expert Answer

Ched1950
Answered 2021-11-16 Author has 1563 answers

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:
\(\displaystyle{\left({x}-{\left(-{4}\right)}\right)}^{{{2}}}+{\left({y}-{\left(-{5}\right)}\right)}^{{{2}}}={\left({\frac{{{9}}}{{\sqrt{{{13}}}}}}\right)}^{{{2}}}\)
\(\displaystyle{\left({x}+{4}\right)}^{{{2}}}+{\left({y}+{5}\right)}^{{{2}}}={\frac{{{81}}}{{{13}}}}\)

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