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

A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of $60^o$. The area enclosed by these tangents and the arc 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 $........$.

If line $ax + by = 0$ touches ${x^2} + {y^2} + 2x + 4y = 0$ and is a normal to the circle ${x^2} + {y^2} - 4x + 2y - 3 = 0$, then value of $(a,b)$ will be

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 :

Tangents are drawn from the point $(4, 3)$ to the circle ${x^2} + {y^2} = 9$. The area of the triangle formed by them and the line joining their points of contact is