For the circle $x^2 + y^2 - 2x + 4y - 4 = 0$,what is the line $2x - y - 1 = 0$?

If the equations $2x - 3y + 3 = 0$,$2x + y + 1 = 0$,and $6x + 4y + 1 = 0$ represent the sides of a triangle,then the equation of the circle passing through the vertices of this triangle is

The equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2+y^2-6x-4y-12=0$ is

If the line $y - 1 = m(x - 1)$ intersects the circle $x^2 + y^2 = 4$ at two distinct points,then the number of possible values of $m$ is:

Let $AB$ be a line segment with mid-point $C$ and $D$ be the mid-point of $AC$. Let $C_1$ be the circle with diameter $AB$ and $C_2$ be the circle with diameter $AC$. Let $E$ be a point on $C_1$ such that $EC$ is perpendicular to $AB$. Let $F$ be a point on $C_2$ such that $DF$ is perpendicular to $AB$ and $E$ and $F$ lie on opposite sides of $AB$. Then,the value of $\sin \angle FEC$ is

Two circles $x^2 + y^2 = ax$ and $x^2 + y^2 = c^2$ touch each other if: