The equation of the first circle is $S_1: x^2 + y^2 - 16x - 20y + 100 = 0$. Its center $C_1 = (-(-16)/2, -(-20)/2) = (8, 10)$ and radius $r_1 = \sqrt{8^2 + 10^2 - 100} = \sqrt{64 + 100 - 100} = \sqrt{64} = 8$. The equation of the second circle is $S_2: (x-4)^2 + (y-7)^2 = 36$. Expanding this gives $x^2 - 8x + 16 + y^2 - 14y + 49 = 36 \Rightarrow x^2 + y^2 - 8x - 14y + 29 = 0$. Its center $C_2 = (4, 7)$ and radius $r_2 = \sqrt{36} = 6$. First, check the condition for intersection: the distance between centers $C_1C_2 = \sqrt{(8-4)^2 + (10-7)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$. Since $|r_1-r_2| = |8-6| = 2$ and $r_1+r_2 = 8+6 = 14$, and $2 < 5 < 14$, the circles intersect at two distinct points. The equation of the common chord of two intersecting circles $S_1=0$ and $S_2=0$ is given by $S_1 - S_2 = 0$. $(x^2 + y^2 - 16x - 20y + 100) - (x^2 + y^2 - 8x - 14y + 29) = 0$. Simplifying the equation: $(-16x - (-8x)) + (-20y - (-14y)) + (100 - 29) = 0$ $-16x + 8x - 20y + 14y + 71 = 0$ $-8x - 6y + 71 = 0$ Multiplying by $-1$ to get positive coefficients for $x$ and $y$: $8x + 6y - 71 = 0$. Thus, option C is correct.