enfurezca3x

2021-11-15

Write the equation of the circle with center at and touching the line $3x-2y-7=0$

Ched1950

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

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