If a circle passes through the point (1, 2) and cuts the circle x2+y2=4 orthogonally, then the equation of the locus of its centre, is

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If a circle passes through the point (1, 2) and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the equation of the locus of its centre, is

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$2 x + 4 y - 9 = 0$

$x^{2} + y^{2} - 3 x - 8 y + 1 = 0$

$2 x + 4 y - 1 = 0$

$x^{2} + y^{2} - 2 x - 6 y - 7 = 0$

answer is C.

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Detailed Solution

Let the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . This passes through $(1, 2) .$

$∴ 5 + 2 g + 4 f + c = 0$ …(i)

The circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ cut orthogonally.

$∴ 2 (g \times 0 + f \times 0) = c - 4 \Rightarrow c = 4$

Putting $c = 4$ in (i), we get $2 g + 4 f + 9 = 0 .$

Therefore, the locus of $(- g, - f)$ is

$- 2 x - 4 y + 9 = 0$ or, $2 x + 4 y - 9 = 0$ .

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