Write the equation of the circle with center at (−4, −5) and touching the line 3x−2y−7=0

Question

Write the equation of the circle with center at (−4, −5) and touching the line 3x−2y−7=0

Ched1950 · Accepted Answer

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

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

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