$ABCD$ is a square with side length $a$. Taking $AB$ and $AD$ as the coordinate axes,find the equation of the circle passing through the vertices of the square.

The equation of a circle which touches the straight lines $x+y=2$ and $x-y=2$ and also touches the circle $x^2+y^2=1$ is:

If the point $(1, 4)$ lies inside the circle $x^2 + y^2 - 6x - 10y + p = 0$ and the circle does not touch or intersect the coordinate axes,then the set of all possible values of $p$ is the interval

Three circles lie on a plane such that each of them externally touches the other two. Two of them have radius $3$,and the third has radius $1$. If $A, B$,and $C$ are the points of tangency of the circles,then the area of the triangle $ABC$ is

$A$ tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. $A$ straight line $L$,perpendicular to $PT$,is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ $A$ common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ $A$ possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

If $P(2, 8)$ is an interior point of a circle $x^2 + y^2 - 2x + 4y - p = 0$ which neither touches nor intersects the axes,then the set for $p$ is