Find the centre and radius of the circle $x^{2} + y^{2} + 2 a x - 2 b y + b^{2} = 0$ .

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Solution

The equation of the circle is given as $x^{2} + y^{2} + 2 a x - 2 b y + b^{2} = 0$ .

Comparing the given equation with general equation of circle,

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

The center of circle is $C = (- g, - f)$ and the radius of circle is $r = \sqrt{g^{2} + f^{2} - c}$ .

Therefore, from the given equation,

$g = a, f = - b$

$C = (- a, b)$

And the radius is,

$r = \sqrt{{(a)}^{2} + {(- b)}^{2} - (b^{2})}$

$= \sqrt{a^{2} + b^{2} - b^{2}}$

$= \sqrt{a^{2}}$

$= a$

Therefore, the center of the circle is $(- a, b)$ and radius of the circle is $a$ .

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