Equation(s) of circle(s) concentric with $x^{2} + y^{2} + 2 x + 4 y = 0$ and touching the circle $x^{2} + y^{2} - 2 x - 1 = 0$ is/are

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$x^{2} + y^{2} + 2 x + 4 y + 3 = 0$

$x^{2} + y^{2} + 2 x + 4 y - 13 = 0$

$x^{2} + y^{2} + 2 x + 4 y + 13 = 0$

$x^{2} + y^{2} + 2 x + 4 y - 3 = 0$

answer is A, B.

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Detailed Solution

Centre of the required circle is (-1, -2), let its radius be ‘r’ it touches $x^{2} + y^{2} - 2 x - 1 = 0,$ (may be internally or externally)
$\Rightarrow$ Distance between centers $= | r \pm \sqrt{2} |$
$\begin{array}{l} r \pm \sqrt{2} = \sqrt{{(1 + 1)}^{2} + {(+ 2)}^{2}} = 2 \sqrt{2} \\ \Rightarrow r = \sqrt{2} \end{array}$
$∴$ Required circles are, ${(x + 1)}^{2} + {(y + 2)}^{2} = 2$
$\begin{array}{l} \Rightarrow x^{2} + y^{2} + 2 x + 4 y + 3 = 0 a n d {(x + 1)}^{2} + {(y + 2)}^{2} = 18 \\ \Rightarrow x^{2} + y^{2} + 2 x + 4 y - 13 = 0 \\ | r - \sqrt{2} | = 2 \sqrt{2} \\ r - \sqrt{2} = 2 \sqrt{2} o r r - \sqrt{2} \neq - 2 \sqrt{2} \\ \Rightarrow r = 3 \sqrt{2} \end{array}$