The equation of the circle having the lines ${x^2} + 2xy + 3x + 6y = 0$ as its normals and having size just sufficient to contain the circle $x(x - 4) + y(y - 3) = 0$is

The value of k so that ${x^2} + {y^2} + kx + 4y + 2 = 0$ and $2({x^2} + {y^2}) - 4x - 3y + k = 0$ cut orthogonally is

Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2 x-$ $6 y+\alpha=0$ in line $y=x+1$ be $c_{2}: 5 x^{2}+5 y^{2}+10 g x$ $+10 f y +38=0$. If $r$ is the radius of circle $c _{2}$, then $\alpha+6 r^{2}$ is equal to$.....$

Consider the circles ${x^2} + {(y - 1)^2} = $ $9,{(x - 1)^2} + {y^2} = 25$. They are such that

One of the limit point of the coaxial system of circles containing ${x^2} + {y^2} - 6x - 6y + 4 = 0$, ${x^2} + {y^2} - 2x$ $ - 4y + 3 = 0$ 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