The circle {x^2 + y^2 + 2gx + 2fy + c = 0} bi | vedclass

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(A) Let the two circles be ${S_1: x^2 + y^2 + 2gx + 2fy + c = 0}$ and ${S_2: x^2 + y^2 + 2g'x + 2f'y + c' = 0}$.If ${S_1}$ bisects the circumference o

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$2g'(g - g') + 2f'(f - f') = c - c'$

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(A) Let the two circles be ${S_1: x^2 + y^2 + 2gx + 2fy + c = 0}$ and ${S_2: x^2 + y^2 + 2g'x + 2f'y + c' = 0}$.If ${S_1}$ bisects the circumference of ${S_2}$,then the common chord of the two circles must pass through the center of ${S_2}$.The equation of the common chord is given by ${S_1 - S_2 = 0}$,which is ${2(g - g')x + 2(f - f')y + (c - c') = 0}$.The center of ${S_2}$ is ${(-g', -f')}$.Substituting the center ${(-g', -f')}$ into the equation of the common chord:${2(g - g')(-g') + 2(f - f')(-f') + (c - c') = 0}$.Rearranging the terms,we get ${2g'(g - g') + 2f'(f - f') = c - c'}$.

The circle ${x^2 + y^2 + 2gx + 2fy + c = 0}$ bisects the circumference of the circle ${x^2 + y^2 + 2g'x + 2f'y + c' = 0}$,if

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