If P is the point of contact of the circles x | vedclass

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(D) Given circles are $S_1: x^2+y^2+4x+4y-10=0$ and $S_2: x^2+y^2-6x-6y+10=0$.For $S_1$,center $C_1 = (-2, -2)$ and radius $r_1 = \sqrt{(-2)^2+(-2)^2-

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$x^2+y^2-14x-14y+26=0$

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(D) Given circles are $S_1: x^2+y^2+4x+4y-10=0$ and $S_2: x^2+y^2-6x-6y+10=0$.For $S_1$,center $C_1 = (-2, -2)$ and radius $r_1 = \sqrt{(-2)^2+(-2)^2-(-10)} = \sqrt{4+4+10} = \sqrt{18} = 3\sqrt{2}$.For $S_2$,center $C_2 = (3, 3)$ and radius $r_2 = \sqrt{3^2+3^2-10} = \sqrt{9+9-10} = \sqrt{8} = 2\sqrt{2}$.The point of contact $P$ divides the line segment $C_1C_2$ internally in the ratio $r_1:r_2 = 3\sqrt{2}:2\sqrt{2} = 3:2$.$P = \left(\frac{3(3)+2(-2)}{3+2}, \frac{3(3)+2(-2)}{3+2}\right) = \left(\frac{9-4}{5}, \frac{9-4}{5}\right) = (1, 1)$.The external centre of similitude $Q$ divides the line segment $C_1C_2$ externally in the ratio $r_1:r_2 = 3:2$.$Q = \left(\frac{3(3)-2(-2)}{3-2}, \frac{3(3)-2(-2)}{3-2}\right) = \left(\frac{9+4}{1}, \frac{9+4}{1}\right) = (13, 13)$.The equation of the circle with $P(1, 1)$ and $Q(13, 13)$ as the extremities of its diameter is $(x-1)(x-13) + (y-1)(y-13) = 0$.$x^2 - 14x + 13 + y^2 - 14y + 13 = 0$.$x^2 + y^2 - 14x - 14y + 26 = 0$.

If $P$ is the point of contact of the circles $x^2+y^2+4x+4y-10=0$ and $x^2+y^2-6x-6y+10=0$,and $Q$ is their external centre of similitude,then the equation of the circle with $P$ and $Q$ as the extremities of its diameter is

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