Find the value of m+n,if the circumference of | vedclass

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(A) Let $S_1 \equiv x^2+y^2+8x+8y-m=0$ and $S_2 \equiv x^2+y^2-2x+4y+n=0$. The common chord of the two circles is given by $S_1 - S_2 = 0$. $(x^2+y^2+

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-$56$

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(A) Let $S_1 \equiv x^2+y^2+8x+8y-m=0$ and $S_2 \equiv x^2+y^2-2x+4y+n=0$. The common chord of the two circles is given by $S_1 - S_2 = 0$. $(x^2+y^2+8x+8y-m) - (x^2+y^2-2x+4y+n) = 0$. $10x + 4y - (m+n) = 0$. Since the circumference of the circle $S_1$ is bisected by the circle $S_2$,the common chord must pass through the center of the circle $S_1$. The center of $S_1$ is $(-4, -4)$. Substituting $(-4, -4)$ into the equation of the common chord: $10(-4) + 4(-4) = m+n$. $-40 - 16 = m+n$. $m+n = -56$.

Find the value of $m+n$,if the circumference of the circle $x^2+y^2+8x+8y-m=0$ is bisected by the circle $x^2+y^2-2x+4y+n=0$.

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