The equation of the tangent to the circle,given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it,is

If the lines $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$ are both tangents to a circle,then its radius is

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6x+4y+8=0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $OPQ$ passes through the point $(\alpha, \frac{1}{2})$,then a value of $\alpha$ is

If the tangent at a point $P$ on the circle $x^2 + y^2 + 6x + 6y = 2$ meets the line $5x - 2y + 6 = 0$ at a point $Q$ on the $y$-axis,then the length of $PQ$ is . . . . .

The gradient of the normal at the point $(-2, -3)$ on the circle $x^2 + y^2 + 2x + 4y + 3 = 0$ is

The angle between the two tangents drawn from the origin to the circle $x^2+y^2-14x+2y+25=0$ is (in $^{\circ}$)