Find the equation of the circle concentric with the circle
$x^{2} + y^{2} - 6 x + 12 y + 15 = 0$
and double of its area.

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Solution

The given equation of circle is
$x^{2} + y^{2} - 6 x + 12 y + 15 = 0$
$∴$ centre $= (- g, - f) = (3, - 6)$
radius= $\sqrt{g^{2} + f^{2} - c} = \sqrt{9 + 36 - 15} = \sqrt{30}$
Now,
the required equation of circle in concentric with (i) which means both have same centre (3. -6)
Also,
Area of required circle = $2 \times$ Area of $π r^{2} = 2 \times π (\sqrt{30})^{2}$
$\Rightarrow R^{2} = 60 \Rightarrow R = 2 \sqrt{15}$
Thus,
the required circle is
$(x - 3)^{2} + (y + 6)^{2} = 60 x^{2} + y^{2} - 6 x + 12 y - 15 = 0$

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