The position of the point $(1, 1)$ with respect to the circle $x^2 + y^2 - x + y - 1 = 0$ is:

The value of $a$,such that the power of the point $(1, 6)$ with respect to the circle $x^2 + y^2 + 4x - 6y - a = 0$ is $-16$,is

$A$ square is inscribed in the circle $x^2 + y^2 - 2x + 4y - 93 = 0$ with its sides parallel to the coordinate axes. The coordinates of its vertices are

$A$ square is inscribed in the circle $x^2+y^2-10x-6y+30=0$. One side of this square is parallel to $y=x+3$. If $(x_i, y_i)$ are the vertices of the square,then $\sum(x_i^2+y_i^2)$ is equal to:

Let $C_1$ and $C_2$ be two circles touching each other externally at point $A$. Let $AB$ be the diameter of circle $C_1$. Draw a secant $BA_3$ to circle $C_2$,intersecting circle $C_1$ at point $A_1$ (where $A_1 \neq A$) and circle $C_2$ at points $A_2$ and $A_3$. If $BA_1 = 2$,$BA_2 = 3$,and $BA_3 = 4$,then the radii of circles $C_1$ and $C_2$ are respectively:

The point diametrically opposite to the point $P(1, 0)$ on the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is