Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$, where $\alpha, \beta>0$. If the circle touches the line $y+x=0$ at the point $P$, whose distance from the origin is $4 \sqrt{2}$ , then $(\alpha+\beta)^2$ is equal to................

If the equation of the tangent to the circle ${x^2} + {y^2} - 2x + 6y - 6 = 0$ parallel to $3x - 4y + 7 = 0$ is $3x - 4y + k = 0$, then the values of $k$ are

The line $(x - a)\cos \alpha + (y - b)$ $\sin \alpha = r$ will be a tangent to the circle ${(x - a)^2} + {(y - b)^2} = {r^2}$

The line $y = mx + c$ will be a normal to the circle with radius $r$ and centre at $(a, b)$, if

If the length of the tangents drawn from the point $(1,2)$ to the circles ${x^2} + {y^2} + x + y - 4 = 0$ and $3{x^2} + 3{y^2} - x - y + k = 0$ be in the ratio $4 : 3$, then $k =$

The equation of three circles are ${x^2} + {y^2} - 12x - 16y + 64 = 0,$ $3{x^2} + 3{y^2} - 36x + 81 = 0$ and ${x^2} + {y^2} - 16x + 81 = 0.$ The co-ordinates of the point from which the length of tangent drawn to each of the three circle is equal is