Given, $C_1: x^2+y^2+30 x-2 y+1=0$
$$
\begin{aligned}
& \therefore \text { centre }(O)=(-15,1) \\
& \text { and radius }=\sqrt{225+1-1}=\sqrt{225}=15 \\
& \text { and } C_2: x^2+y^2-14 x+6 y+33=0 \\
& \therefore \text { Centre }\left(O^{\prime}\right)=(7,-3) \\
& \text { and radius }=\sqrt{49+9-33}=\sqrt{25}=5
\end{aligned}
$$

Since, point $T$ divides $O O^{\prime}$ in $15: 5$ i.e., $3: 1$ internally.
$$
\therefore \quad T=\left(\frac{21-15}{4}, \frac{-9+1}{4}\right)=\left(\frac{3}{2},-2\right)
$$

Also, point $D$ divides $O O^{\prime}$ in $15: 5$ i.e., $3: 1$ externally.
$$
D=\left(\frac{21+15}{2}, \frac{-9-1}{2}\right)=(18,-5)
$$

Now, centre of circle with $T D$ as diameters of mid-point of $T D$
$$
=\left(\frac{18+3 / 2}{2}, \frac{-2-5}{2}\right)=\left(\frac{39}{4}, \frac{-7}{2}\right)
$$