The equations of the tangents to the circle ${x^2} + {y^2} - 6x + 4y = 12$ which are parallel to the straight line $4x + 3y + 5 = 0$, are
$3x - 4y - 19 = 0,\,\,3x - 4y + 31 = 0$
$4x + 3y - 19 = 0,\,\,4x + 3y + 31 = 0$
$4x + 3y + 19 = 0,\,\,4x + 3y - 31 = 0$
$3x - 4y + 19 = 0,3x - 4y + 31 = 0$
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 gradient of the tangent line at the point $(a\cos \alpha ,a\sin \alpha )$ to the circle ${x^2} + {y^2} = {a^2}$, is
The point at which the normal to the circle ${x^2} + {y^2} + 4x + 6y - 39 = 0$ at the point $(2, 3)$ will meet the circle again, is
If the point $(1, 4)$ lies inside the circle $x^2 + y^2-6x - 10y + p = 0$ and the circle does not touch or intersect the coordinate axes, then the set of all possible values of $p$ is the interval
The equations of the normals to the circle ${x^2} + {y^2} - 8x - 2y + 12 = 0$ at the points whose ordinate is $-1,$ will be