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The locus of centres of all the circles passing through $(1,2)$ and cutting $\mathrm{x^2+y^2=16}$ orthogonally is

Option: 1
$\mathrm{4 x-2 y+21=0}$

Option: 2
$\mathrm{x+2 y+11=0}$

Option: 3
$\mathrm{2 x+4 y-21=0}$

Option: 4
$\mathrm{2 x-y+11=0}$

Answers (1)

best_answer

$\mathrm{ x^2+y^2+2 g x+2 f y+c=0 }$

$\mathrm{ x^2+y^2-16=0}$

circle cutting orthogonally

so that, $\mathrm{2 g(0)+2 f(0)=c-16 \Rightarrow c=16}$
Now the circle passes through $\mathrm{(1,2)}$ , so that

$\mathrm{ 1+4+2 g+4 f+c=0 }$

$\mathrm{ c=-2 g-4 f-5 }$

$\mathrm{ -2 g-4 f-5=16 }$

$\mathrm{ 2 g+4 f+21=0 }$ $\mathrm{ \Rightarrow Required\: locus\: of (-g,-f) is }$

$\mathrm{ 2 x+4 y-21=0}$

Hence option 3 is correct.

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manish painkra

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