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If the circle \mathrm{x^2+y^2=4} bisects the circumference of the circle \mathrm{x^2+y^2-2 x+6 y+a=0}, then a equals

Option: 1

4


Option: 2

-4


Option: 3

16


Option: 4

-16


Answers (1)

best_answer

The common chord of given circles is

\mathrm{ S_1-S_2=0 \Rightarrow 2 x-6 y-4-a=0 }         \mathrm{.....(i)}

Since, \mathrm{x^2+y^2=4} bisects the circumferences of the circle

\mathrm{x^2+y^2-2 x+6 y+a=0},

therefore (i) passes through the centre of second circle i.e.

\mathrm{(1,-3) . \quad \therefore \quad 2+18-4-a=0 \Rightarrow a=16}

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Ritika Harsh

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