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(D) For the circle $S_1 \equiv x^2+y^2-14x+6y+33=0$,the center $C_1 = (7, -3)$ and radius $r_1 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49+9-33} = 5$. For t

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$\left(\frac{39}{4}, \frac{-7}{2}\right)$

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(D) For the circle $S_1 \equiv x^2+y^2-14x+6y+33=0$,the center $C_1 = (7, -3)$ and radius $r_1 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49+9-33} = 5$. For the circle $S_2 \equiv x^2+y^2+30x-2y+1=0$,the center $C_2 = (-15, 1)$ and radius $r_2 = \sqrt{(-15)^2 + 1^2 - 1} = \sqrt{225+1-1} = 15$. The internal center of similitude $A$ divides $C_1C_2$ in the ratio $r_1:r_2 = 5:15 = 1:3$. $A = \left(\frac{1(-15) + 3(7)}{1+3}, \frac{1(1) + 3(-3)}{1+3}\right) = \left(\frac{-15+21}{4}, \frac{1-9}{4}\right) = \left(\frac{6}{4}, \frac{-8}{4}\right) = \left(\frac{3}{2}, -2\right)$. The external center of similitude $B$ divides $C_1C_2$ in the ratio $r_1:r_2 = 1:3$ externally. $B = \left(\frac{1(-15) - 3(7)}{1-3}, \frac{1(1) - 3(-3)}{1-3}\right) = \left(\frac{-15-21}{-2}, \frac{1+9}{-2}\right) = \left(\frac{-36}{-2}, \frac{10}{-2}\right) = (18, -5)$. The midpoint of $AB$ is $\left(\frac{\frac{3}{2} + 18}{2}, \frac{-2 - 5}{2}\right) = \left(\frac{\frac{39}{2}}{2}, \frac{-7}{2}\right) = \left(\frac{39}{4}, \frac{-7}{2}\right)$.

If $A$ and $B$ are the centres of similitude with respect to the circles $x^2+y^2-14x+6y+33=0$ and $x^2+y^2+30x-2y+1=0$,then the midpoint of $AB$ is

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