The equation of the chord of the circle $x^2+y^2-4x=0$ whose midpoint is $(1,0)$ is

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:

Let the line $x - y = 4$ intersect the circle $C : (x - 4)^2 + (y + 3)^2 = 9$ at the points $Q$ and $R$. If $P(\alpha, \beta)$ is a point on $C$ such that $PQ = PR$, then $(6\alpha + 8\beta)^2$ is equal to . . . . . .

If ${c^2} > {a^2}(1 + {m^2})$,then the line $y = mx + c$ will intersect the circle ${x^2} + {y^2} = {a^2}$ at:

If a direct common tangent drawn to the circles $x^2+y^2-6x+4y+9=0$ and $x^2+y^2+2x-2y+1=0$ touches the circles at $A$ and $B$,then $AB=$

If $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle $x^2+y^2=25$,then $4a+2b=$