A tangent drawn from the point $(4, 0)$ to the circle $x^2 + y^2 = 8$ touches it at a point $A$ in the first quadrant. The co-ordinates of another point $B$ on the circle such that $l\, (AB) = 4$ are :

Let $PQ$ and $RS$ be the tangent at the extremities of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $(PQ.RS)$ is equal to

The area of the triangle formed by the positive $x$-axis and the normal and the tangent to the circle $x^2 + y^2 = 4$ at $(1, \sqrt 3 )$ is

The angle of intersection of the circles ${x^2} + {y^2} - x + y - 8 = 0$ and ${x^2} + {y^2} + 2x + 2y - 11 = 0,$ is

A line meets the co-ordinate axes in $A\, \& \,B. \,A$ circle is circumscribed about the triangle $OAB.$ If $d_1\, \& \,d_2$ are the distances of the tangent to the circle at the origin $O$ from the points $A$ and $B$ respectively, the diameter of the circle is :

The equation of the tangent to the circle ${x^2} + {y^2} = {r^2}$ at $(a,b)$ is $ax + by - \lambda = 0$, where $\lambda $ is