Suppose we have two circles of radius $2$ each in the plane such that the distance between their centers is $2 \sqrt{3}$. The area of the region common to both circles lies between

Let the circles $C_1: (x-\alpha)^2 + (y-\beta)^2 = r_1^2$ and $C_2: (x-8)^2 + (y-\frac{15}{2})^2 = r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2:1$,then $(\alpha+\beta) + 4(r_1^2 + r_2^2)$ equals

If ${c^2} > {a^2}(1 + {m^2})$,then the line $y = mx + c$ will intersect the circle ${x^2} + {y^2} = {a^2}$ at:

Let $y=x+2$,$4y=3x+6$,and $3y=4x+1$ be three tangent lines to the circle $(x-h)^2+(y-k)^2=r^2$. Then $h+k$ is equal to:

When is the angle of intersection of two circles $0^{\circ}$?

Let a triangle $ABC$ be inscribed in the circle $x^{2} - \sqrt{2}(x+y) + y^{2} = 0$ such that $\angle BAC = \frac{\pi}{2}$. If the length of side $AB$ is $\sqrt{2}$,then the area of the $\triangle ABC$ is equal to