Find the equation of the chord of the circle $x^2 + y^2 = 16$ which is bisected at the point $(2, -1)$.

If $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$ are two tangents to a circle,then the radius of the circle is

Let a circle $C$ touch the lines $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$,where $K_{1}, K_{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(-1, 2)$ and $L_{2}$ at $(3, -6)$,then the equation of the circle $C$ is:

The mid-point of the chord of the circle $x^2+y^2-6x+4y-12=0$ drawn parallel to the tangent at $(-1,1)$ and at a distance of one unit from the tangent 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 $C_1$ externally at the point $(\alpha, \beta)$. If $(\beta-\alpha)^2=\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m + n$ is equal to :

Let a circle $C_1 \equiv x^2 + y^2 - 4x + 6y + 1 = 0$ and circle $C_2$ be such that its centre is the image of the centre of $C_1$ about the $x$-axis and the radius of $C_2$ is equal to the radius of $C_1$. Then,the area of $C_1$ which is not common with $C_2$ is: