If $C_1$ and $C_2$ are the centres of similitude with respect to the circles $x^2+y^2-14 x+6 y+33=0$ and $x^2+y^2+30 x-2 y+1=0$, then the equation of the circle with $C_1 C_2$ as diameter is

Given equation of circles are $$ x^2+y^2-14 x+6 y+33=0 $$ and $$ x^2+y^2+30 x-2 y+1=0 $$ Clearly, circle (i) has centre $O_1(7,-3)$ $$ \text { and radius } r_1=\sqrt{49+9-33}=5 $$ and circle (ii) has centre $O_2(-15,1)$ and radius $r_2=\sqrt{225+1-1}=15$ We know that the centres of similitude of two circles, whose centres are $A$ and $B$, are the points which divide $A B$ internally and externally in the ratio of radii $r_a, r_b$. So, $C_1=$ point which divide $O_1 O_2$, internally in the ratio $5: 15$, i.e. $1: 3$ $$ \begin{aligned} & =\left(\frac{-15+21}{4}, \frac{1-9}{4}\right)=\left(\frac{3}{2},-2\right) \\ & \frac{1 \vdots 3}{O_1(7,-3) \quad} \quad O_2(-15,1) \end{aligned} $$ and $C_2=$ Point which divide $O_1 O_2$ externally in the ratio $1: 3$
$$ =\left(\frac{-15-21}{1-3}, \frac{1+9}{1-3}\right)=\left(\frac{-36}{-2}, \frac{10}{-2}\right)=(18,-5) $$ Now, equation of circle with as $C_1 C_2$ diameter is given by $$ \begin{aligned} \left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right) & =0 \\ =\left(x-\frac{3}{2}\right)(x-18)+(y+2)(y+5) & =0 \\ \Rightarrow \quad(2 x-3)(x-18)+2(y+4)(y+5) & =0 \\ \left.\Rightarrow 2 x^2-36 x-3 x+54+2 y^2+7 y+10\right) & =0 \\ \Rightarrow \quad 2 x^2-39 x+54+2 y^2+14 y+20 & =0 \\ \Rightarrow \quad 2 x^2+2 y^2-39 x+14 y+74 & =0 \end{aligned} $$