The equation of the circle circumscribing the triangle formed by the straight lines $x+y=6$,$2x+y=4$,and $x+2y=5$ is given by:

The curve $xy = c, (c > 0)$,and the circle $x^2 + y^2 = 1$ touch at two points. Then the distance between the points of contact is

$A$ circle passing through the point $P(\alpha, \beta)$ in the first quadrant touches the two coordinate axes at the points $A$ and $B$. The point $P$ is above the line $AB$. The point $Q$ on the line segment $AB$ is the foot of the perpendicular from $P$ on $AB$. If $PQ$ is equal to $11$ units,then the value of $\alpha \beta$ is $.............$.

The angle between the circles $x^2+y^2-2x-9=0$ and $x^2+y^2-4y-1=0$ at their point of intersection is

The inverse point of $(1, 3)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $PA$ meets the circle in $Q$ with $PQ=5$,then $c$ is equal to