If the circle x^2 + y^2 = 4 bisects the circu | vedclass

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(C) The equation of the first circle is $S_1: x^2 + y^2 - 4 = 0$.The equation of the second circle is $S_2: x^2 + y^2 - 2x + 6y + a = 0$.The common ch

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$16$

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(C) The equation of the first circle is $S_1: x^2 + y^2 - 4 = 0$.The equation of the second circle is $S_2: x^2 + y^2 - 2x + 6y + a = 0$.The common chord of the two circles is given by $S_1 - S_2 = 0$,which is $(x^2 + y^2 - 4) - (x^2 + y^2 - 2x + 6y + a) = 0$.This simplifies to $2x - 6y - 4 - a = 0$.If a circle bisects the circumference of another circle,the common chord must be a diameter of the second circle.Therefore,the common chord must pass through the center of the second circle,which is $(1, -3)$.Substituting $(1, -3)$ into the equation of the common chord: $2(1) - 6(-3) - 4 - a = 0$.$2 + 18 - 4 - a = 0$.$16 - a = 0$,which gives $a = 16$.

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

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