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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Solution

The correct option is C. $2 x + 4 y - 9 = 0$
Let the equation of the circle by $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Since, the circle passes through $(1, 2)$ , therefore,
$\Rightarrow 1^{2} + 2^{2} + 2 g x + 2 f y + c = 0$
$\Rightarrow 5 + 2 g + 4 f + c = 0$ ...(i)
Given equation of the circle is $x^{2} + y^{2} = 4$ ...(ii)
On comparing with the general equation of circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , we get,
$g_{1} = - 1, g_{2} = 0, f_{1} = - 2, f_{2} = 0, c_{1} = c$ and $c_{2} = - 4$
Condition of orthogonality is $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
$\Rightarrow 0 + 0 = c - 4$
$\Rightarrow c = 4$
On putting value in (i), we get,
$\Rightarrow 2 g + 4 f + 9 = 0$
Hence, locus of the centre $- g, - f$ is $- 2 x - 4 y + 9 = 0$ or $2 x + 4 y - 9 = 0$

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