The number of real circles cutting orthogonally the circle $x^{2}+y^{2}+2x-2y+7=0$ is

$ABCD$ 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.

For the given circle $2x^2 + 2y^2 = 5$ and the parabola $y^2 = 4\sqrt{5}x$:
Statement-$I$: The equation of the common tangent to these curves is $y = x + \sqrt{5}$.
Statement-$II$: If the line $y = mx + \frac{\sqrt{5}}{m} (m \neq 0)$ is a common tangent,then $m$ satisfies $m^4 - 3m^2 + 2 = 0$.

If $\theta$ is the angle made by the common tangent to the circle $x^2+y^2=16$ and the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ with the positive $X$-axis,then $\cos 2 \theta=$

$A$ line drawn from a fixed point $P(\alpha, \beta)$ intersects the circle $x^2 + y^2 = r^2$ at points $A$ and $B$. Then $PA \cdot PB = \dots$

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: