Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to:

The centre of the circle that passes through the point $(0,1)$ and touches the curve $y=x^2$ at $(2,4)$ is

For each natural number $k$,let $C_k$ denote the circle with radius $k$ centimeters and center at the origin. On the circle $C_k$,a particle moves $k$ centimeters in the counter-clockwise direction. After completing its motion on $C_k$,the particle moves to $C_{k+1}$ in the radial direction. The motion of the particle continues in this manner. The particle starts at $(1, 0)$. If the particle crosses the positive direction of the $x$-axis for the first time on the circle $C_n$,then $n$ is equal to

For the circle $x^2 + y^2 - 2x + 4y - 4 = 0$,what is the line $2x - y - 1 = 0$?

If the circles $(x+a)^2+(y+b)^2=a^2$ and $(x+c)^2+(y+d)^2=d^2$ cut orthogonally,then $b(b-2d) =$

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