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(C) Given circles are $C_1: x^2+y^2+30x-2y+1=0$ and $C_2: x^2+y^2-14x+6y+33=0$.For $C_1$,centre $O = (-15, 1)$ and radius $r_1 = \sqrt{(-15)^2 + 1^2 -

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

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(C) Given circles are $C_1: x^2+y^2+30x-2y+1=0$ and $C_2: x^2+y^2-14x+6y+33=0$.For $C_1$,centre $O = (-15, 1)$ and radius $r_1 = \sqrt{(-15)^2 + 1^2 - 1} = \sqrt{225} = 15$.For $C_2$,centre $O' = (7, -3)$ and radius $r_2 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49+9-33} = \sqrt{25} = 5$.The point $T$ is the intersection of transverse common tangents,which divides the line segment joining the centres $O$ and $O'$ internally in the ratio $r_1 : r_2 = 15 : 5 = 3 : 1$.$T = \left(\frac{3(7) + 1(-15)}{3+1}, \frac{3(-3) + 1(1)}{3+1}\right) = \left(\frac{21-15}{4}, \frac{-9+1}{4}\right) = \left(\frac{6}{4}, \frac{-8}{4}\right) = \left(\frac{3}{2}, -2\right)$.The point $D$ is the intersection of direct common tangents,which divides the line segment joining the centres $O$ and $O'$ externally in the ratio $r_1 : r_2 = 3 : 1$.$D = \left(\frac{3(7) - 1(-15)}{3-1}, \frac{3(-3) - 1(1)}{3-1}\right) = \left(\frac{21+15}{2}, \frac{-9-1}{2}\right) = \left(\frac{36}{2}, \frac{-10}{2}\right) = (18, -5)$.The centre of the circle having $TD$ as diameter is the midpoint of $TD$.Midpoint $= \left(\frac{3/2 + 18}{2}, \frac{-2 + (-5)}{2}\right) = \left(\frac{3+36}{4}, \frac{-7}{2}\right) = \left(\frac{39}{4}, \frac{-7}{2}\right)$.

If the point of intersection of the pair of the transverse common tangents and that of the pair of direct common tangents drawn to the circles $x^2+y^2-14x+6y+33=0$ and $x^2+y^2+30x-2y+1=0$ are $T$ and $D$ respectively,then the centre of the circle having $TD$ as diameter is

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