If the circle $x^{2} + y^{2} - 6 x - 2 y + 9 = 0$ is completely contained in the circle $x^{2} + y^{2} - 2 x - 5 y + k = 0$ , then the minimum integral value of $| k |$ is

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Solution

$x^{2} + y^{2} - 2 x - 5 y + k = 0 \dots (1)$
Center is $C_{1} (1, \frac{5}{2})$
Radius is $r_{1} = \sqrt{1 + \frac{25}{4} - k} = \sqrt{\frac{29}{4} - k}$

$x^{2} + y^{2} - 6 x - 2 y + 9 = 0 \dots (2)$
Center is $C_{2} (3, 1)$ and radius is $r_{2} = 1$

Circle (2) is completely contained in circle (1)
So, $C_{1} C_{2} \leq r_{1} - r_{2}$
$\Rightarrow \sqrt{4 + \frac{9}{4}} \leq \sqrt{\frac{29}{4} - k} - 1 \Rightarrow (\frac{5}{2} + 1)^{2} \leq \frac{29}{4} - k \Rightarrow k \leq - 5$

Hence, the minimum integral value of $| k | = 5$

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