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 $cm$)

$A$ circle is inscribed in an equilateral triangle of side length $12$. If the area and perimeter of any square inscribed in this circle are $m$ and $n$,respectively,then $m+n^2$ is equal to

Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to:

The equation of the circle which passes through $(1, 0)$ and $(0, 1)$ and has its radius as small as possible,is

The centre of the smallest circle touching the circles $x^2 + y^2 - 2y - 3 = 0$ and $x^2 + y^2 - 8x - 18y + 93 = 0$ is:

If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^2+y^2+4x-6y=12$ externally at the point $(1, -1)$,then the radius of $C$ is