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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2x + 4y - 9 = 0

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2x + 4y - 1 = 0

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Solution

The correct option is C
2x + 4y - 9 = 0

Let equation of circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

It passes through (1, 2) i.e. 1 + 4 + 2g + 4f + c = 0 and is orthogonal w.r.t $x^{2} + y^{2} = 4.$

Therefore, 2g × 0 + 2f × 0 = -c + 4 or c = 4

Hence, we get 2g + 4f + 9 = 0,

then 2x + 4y - 9 = 0 is the required locus.

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