Let a circle of radius 4 pass through the origin O , the points $\mathrm{A}(-\sqrt{3} a, 0)$ and $\mathrm{B}(0,-\sqrt{2} b)$, where $a$ and $b$ are real parameters and $a b \neq 0$. Then the locus of the centroid of $\triangle \mathrm{OAB}$ is a circle of radius

Let the set of all values of $r$, for which the circles $(x+1)^2+(y+4)^2=r^2$ and $x^2+y^2-4 x-2 y-4=0$ intersect at two distinct points be the interval $(\alpha, \beta)$. Then $\alpha \beta$ is equal to

Let PQ and MN be two straight lines touching the circle $x^2+y^2-4 x-6 y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle A O B=\pi / 3$. Then the locus of the point of intersection of the lines PQ and MN is :

Let $C_1$ be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let $C_2$ be the circle with centre $(1,3)$ that touches $\mathrm{C}_1$ externally at the point $(\alpha, \beta)$. If $(\beta-\alpha)^2=\frac{m}{n}$ , $\operatorname{gcd}(m, n)=1$, then $m+n$ is equal to