Given the circles ${x^2} + {y^2} - 4x - 5 = 0$and ${x^2} + {y^2} + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha ,\beta )$such that the tangents from P to both the circles are equal, then

The line $ax + by + c = 0$ is a normal to the circle ${x^2} + {y^2} = {r^2}$. The portion of the line $ax + by + c = 0$ intercepted by this circle is of length

Consider the following statements :

Assertion $(A)$ : The circle ${x^2} + {y^2} = 1$ has exactly two tangents parallel to the $x$ - axis

Reason $(R)$ : $\frac{{dy}}{{dx}} = 0$ on the circle exactly at the point $(0, \pm 1)$.

Of these statements

The centre of the circle passing through the point $(0,1)$ and touching the parabola $y=x^{2}$ at the point $(2,4)$ is

If the equation of one tangent to the circle with centre at $(2, -1)$ from the origin is $3x + y = 0$, then the equation of the other tangent through the origin is

The tangent$(s)$ from the point of intersection of the lines $2x -3y + 1$ = $0$ and $3x -2y -1$ = $0$ to circle $x^2 + y^2 + 2x -4y$ = $0$ will be -