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

Let $O$ be the centre of the circle $x ^2+ y ^2= r ^2$, where $r >\frac{\sqrt{5}}{2}$. Suppose $P Q$ is a chord of this circle and the equation of the line passing through $P$ and $Q$ is $2 x+4 y=5$. If the centre of the circumcircle of the triangle $O P Q$ lies on the line $x+2 y=4$, then the value of $r$ is. . . .

If the line $3x -4y -k = 0 (k > 0)$ touches the circle $x^2 + y^2 -4x -8y -5 = 0$ at $(a, b)$ then $k + a + b$ is equal to :-

The equation of the chord of the circle ${x^2} + {y^2} = {a^2}$ having $({x_1},{y_1})$ as its mid-point is

A circle passes through the points $(- 1, 1) , (0, 6)$ and $(5, 5)$ . The point$(s)$ on this circle, the tangent$(s)$ at which is/are parallel to the straight line joining the origin to its centre is/are :

Let $B$ be the centre of the circle $x^{2}+y^{2}-2 x+4 y+1=0$ Let the tangents at two points $\mathrm{P}$ and $\mathrm{Q}$ on the circle intersect at the point $\mathrm{A}(3,1)$. Then $8.$ $\left(\frac{\text { area } \triangle \mathrm{APQ}}{\text { area } \triangle \mathrm{BPQ}}\right)$ is equal to .... .