If C_1 and C_2 are the centres of similitude | vedclass

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(C) Given equations of the circles are:$x^2+y^2-14 x+6 y+33=0$  $(i)$$x^2+y^2+30 x-2 y+1=0$  $(ii)$For circle $(i)$,the centre is $O_1(7, -3)$ and the

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$2 x^2+2 y^2-39 x+14 y+74=0$

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(C) Given equations of the circles are:$x^2+y^2-14 x+6 y+33=0$  $(i)$$x^2+y^2+30 x-2 y+1=0$  $(ii)$For circle $(i)$,the centre is $O_1(7, -3)$ and the radius is $r_1 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49+9-33} = \sqrt{25} = 5$.For circle $(ii)$,the centre is $O_2(-15, 1)$ and the radius is $r_2 = \sqrt{(-15)^2 + 1^2 - 1} = \sqrt{225+1-1} = \sqrt{225} = 15$.The centres of similitude $C_1$ and $C_2$ divide the line segment joining the centres $O_1$ and $O_2$ internally and externally in the ratio of their radii $r_1 : r_2 = 5 : 15 = 1 : 3$.Internal division point $C_1 = \left( \frac{1(-15) + 3(7)}{1+3}, \frac{1(1) + 3(-3)}{1+3} \right) = \left( \frac{-15+21}{4}, \frac{1-9}{4} \right) = \left( \frac{6}{4}, \frac{-8}{4} \right) = \left( \frac{3}{2}, -2 \right)$.External division point $C_2 = \left( \frac{1(-15) - 3(7)}{1-3}, \frac{1(1) - 3(-3)}{1-3} \right) = \left( \frac{-15-21}{-2}, \frac{1+9}{-2} \right) = \left( \frac{-36}{-2}, \frac{10}{-2} \right) = (18, -5)$.The equation of the circle with $C_1 C_2$ as diameter is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the coordinates,we get $(x - \frac{3}{2})(x - 18) + (y + 2)(y + 5) = 0$.Multiplying by $2$,we get $(2x - 3)(x - 18) + 2(y^2 + 7y + 10) = 0$.$2x^2 - 36x - 3x + 54 + 2y^2 + 14y + 20 = 0$.$2x^2 + 2y^2 - 39x + 14y + 74 = 0$.

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

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