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If two circles ${(x - 1)}^{2} + {(y - 3)}^{2} = r^{2} and x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect at two distinct points then

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$r > 2$

$2 < r < 8$

$r < 2$

$r = 2$

answer is B.

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Detailed Solution

${(x - 1)}^{2} + {(y - 3)}^{2} = r^{2}$

Centre $C_{1} (1, 3)$ and radius $r_{1} = r$

$x^{2} + y^{2} - 8 x + 2 y + 8 = 0$

Centre $C_{2} (4, - 1)$ and radius $r_{2} = \sqrt{4^{2} + {(- 1)}^{2} - 8}$

$= \sqrt{9} = 3$

$C_{1} C_{2} < r_{1} + r_{2}$ and $| r_{1} - r_{2} | < C_{1} C_{2}$

$\begin{array}{l} \sqrt{{(4 - 1)}^{2} + {(- 1 - 3)}^{2}} < r + 3, | r - 3 | < 5 \\ \sqrt{3^{2} + {(- 4)}^{2}} < r + 3, - 5 < r - 3 < 5 \\ 5 < r + 3, - 2 < r < 8 \\ r + 3 > 5 \\ r > 2 \end{array}$

Common solution $2 < r < 8$ .

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