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

Option: 1

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


Option: 2

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


Option: 3

2 x+4 y-9=0


Option: 4

2 x+4 y-1=0


Answers (1)

best_answer

Let the circle be \mathrm{x^{2}+y^{2}+2 g x+2 f y+c=0}. This passes through (1,2) ,therefore 5+2 \mathrm{~g}+4 \mathrm{f}+\mathrm{c}=0\quad \ldots(i)

The circle \mathrm{x^{2}+y^{2}=4} intersects the circle \mathrm{x^{2}+y^{2}+2 g x+2 f y+c=0} orthogonally. Therefore \mathrm{2( g. 0+ f. 0)=c-4 \Rightarrow c=4}

Putting \mathrm{c=-4\, in\: (i), } we get \mathrm{2 g+4 f+9=0}. Therefore, the locus of \mathrm{(-g,-f)} is \mathrm{-2 x-4 y+9=0}, or \mathrm{2 \mathrm{x}+4 \mathrm{y}-9=0}.

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