The equation of the circle whose centre and radius are $(1, - 1)$ and $4$ respectively, is

x^{2} - y^{2} + 2 x + 2 y - 14 = 0

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x^{2} + y^{2} - 2 x - 2 y - 14 = 0

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x^{2} + y^{2} - 2 x + 2 y - 14 = 0

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x^{2} + y^{2} - 2 x - 2 y + 14 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 2 x + 2 y - 14 = 0$
We know that equation of circle is
${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$
Here, centre is $(1, - 1)$ and radius is $4$ .
$∴$ Equation is
${(x - 1)}^{2} + {(y + 1)}^{2} = {(4)}^{2}$
$\Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 + 2 y = 16$
$\Rightarrow x^{2} + y^{2} - 2 x + 2 y - 14 = 0$

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x^{2} + y^{2} + 2 x - 2 y - 2 = 0

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