The equation of a circle with centre at $(2, - 3)$ and the circumference is $10 π$ units is

x^{2} + y^{2} + 4 x + 6 y + 12 = 0

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x^{2} + y^{2} - 4 x + 6 y - 12 = 0

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x^{2} + y^{2} - 4 x + 6 y + 12 = 0

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x^{2} + y^{2} - 4 x - 6 y - 12 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$
For a circle with center (a,b) and radius r,
Equation is $(x - a)^{2} + (y - b)^{2} = r^{2}$

First let's find radius of circle,
$\to$ Circumference = $2 \times π \times r$
$\to$ $10 \times π$ = $2 \times π \times r$
$\to$ $r = 5$

Equation of circle,
$\to$ $(x - 2)^{2} + (y + 3)^{2} = 5^{2}$
$\to$ $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$

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