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The external centre of similitude of the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 2 x - 6 y + 9 = 0$ is

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An Intiative by Sri Chaitanya

(1, 3)

(2, 6)

(3, 9)

(1, 2)

answer is B.

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Detailed Solution

$r_{1} : r_{2} = 2 : 1$

$E x t e r n a l c e n t r e o f s i m i l i t u d e = (\frac{m x_{2} - n x_{1}}{m - n}, \frac{m y_{2} - n y_{1}}{m - n})$

$=$ $\frac{2 (1, 3) - 1 (0, 0)}{1} = (2, 6)$

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