If $y+c=0$ is a tangent to the circle $x^2+y^2-6x-2y+1=0$ at $(a, 4)$,then

Two tangents to the circle $x^2+y^2=4$ at the points $A$ and $B$ meet at the point $P(-4,0)$. Then the area of the quadrilateral $PAOB$,where $O$ is the origin,is:

The line $ax + by + c = 0$ is a normal to the circle $x^2 + y^2 = r^2$. The length of the intercept made by the circle on the line $ax + by + c = 0$ is:

The straight line $x+2y=1$ cuts the $X$-axis at $A$ and the $Y$-axis at $B$. $A$ circle is drawn through $A, B$ and the origin $O(0,0)$. The sum of the perpendicular distances from $A$ and $B$ to the tangent drawn at the origin to the circle $S$ is:

The equation of the tangent to the circle $x^2 + y^2 - 2x - 4y - 4 = 0$ which is perpendicular to $3x - 4y - 1 = 0$ is:

The line $2x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on the line $x - 2y = 4$. Then,the radius of the circle is