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

Tangent to the circle $x^2 + y^2$ = $5$ at the point $(1, -2)$ also touches the circle $x^2 + y^2 -8x + 6y + 20$ = $0$ . Then its point of contact is 

At which point on $y$-axis the line $x = 0$ is a tangent to circle ${x^2} + {y^2} - 2x - 6y + 9 = 0$

Equation of the tangent to the circle ${x^2} + {y^2} = {a^2}$ which is perpendicular to the straight line $y = mx + c$  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 :

A circle with centre $(2,3)$ and radius $4$ intersects the line $x + y =3$ at the points $P$ and $Q$. If the tangents at $P$ and $Q$ intersect at the point $S(\alpha, \beta)$, then $4 \alpha-7 \beta$ is equal to $........$.