The locus of centre of the circle which cuts the circles${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ orthogonally is

Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals

If the two circles, $x^2 + y^2 + 2 g_1x + 2 f_1y = 0\, \& \,x^2 + y^2 + 2 g_2x + 2 f_2y = 0$ touch each then:

Two given circles ${x^2} + {y^2} + ax + by + c = 0$ and ${x^2} + {y^2} + dx + ey + f = 0$ will intersect each other orthogonally, only when

Three circles of radii $a, b, c\, ( a < b < c )$ touch each other externally. If they have $x -$ axis as a common tangent, then