The number of common tangents to two circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 8x + 12 = 0$ is

Chords $AB$ and $CD$ of a circle intersect at a right angle at point $P$. If the lengths of $AP$, $PB$, $CP$, and $PD$ are $2$, $6$, $3$, and $4$ units respectively, then the radius of the circle is:

Two circles of radii $4 \text{ cm}$ and $1 \text{ cm}$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $\sin \theta =$

The points $A(a, 0), B(0, b), C(c, 0)$ and $D(0, d)$ are such that $ac = bd$ and $a, b, c, d$ are all non-zero. Then the points:

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

If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}$,then the value of $r^{2}$ is equal to