The line $2x - y + 1 = 0$ is tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2y = 4$. The radius of the circle is

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k-1)y + 3 = 0$ and $2x + ky - 4 = 0$. If the line $x - y + 2 = 0$ intersects the circle at the points $A$ and $B$, then $(AB)^2$ is equal to:

$B$ and $C$ are fixed points having coordinates $(3, 0)$ and $(-3, 0)$ respectively. If the vertical angle $\angle BAC$ is $90^o$,then the locus of the centroid of the $\Delta ABC$ has the equation:

$A$ circle $C$ of radius $2$ lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has its centre at the point $(2, 5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of $r$ is the interval $(\alpha, \beta)$,then $3 \beta - 2 \alpha$ is equal to:

If the equation of the circumcircle of the triangle formed by the lines $L_1 \equiv x+y=0$,$L_2 \equiv 2x+y-1=0$,and $L_3 \equiv x-3y+2=0$ is $\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L_3 L_1 = 0$,then find the value of $\frac{7 \lambda_1}{\lambda_2} + \frac{\lambda_3}{\lambda_1}$.

For different real non-zero numbers $x_1, x_2, x_3$ and $x_4$,suppose the points $(x_1, \frac{1}{x_1}), (x_2, \frac{1}{x_2}), (x_3, \frac{1}{x_3})$ and $(x_4, \frac{1}{x_4})$ lie on the boundary of a circle of radius $4$. Then,the value of $x_1 x_2 x_3 x_4$ is