Find the radius of the circle in which a central angle of $60^{\circ}$ intercepts an arc of length $37.4 \, cm$ (use $\pi = \frac{22}{7}$). (in $cm$)

If the circle $(x - h)^2 + (y - k)^2 = r^2$ touches the curve $y = x^2 + 1$ at the point $(1, 2)$,then the possible locations of the points $(h, k)$ are given by

The number of common tangents to the circles $x^2 + y^2 - 2x + 4y - 4 = 0$ and $x^2 + y^2 - 8x - 4y + 16 = 0$ is:

The perimeter of a certain sector of a circle is equal to the length of the arc of the semicircle. Then,the angle at the centre of the sector in radians is

The least distance of the point $(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ is

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