If $P$ is a point in the exterior of the circle,then maximum $\ldots \ldots \ldots \ldots$ tangents can be drawn to a circle from $P.$

In the figure,from an external point $P$,a tangent $PT$ and a line segment $PAB$ are drawn to a circle with centre $O$. $ON$ is perpendicular to the chord $AB$. Prove that:
$(i) \quad PA \cdot PB = PN^2 - AN^2$
$(ii) \quad PN^2 - AN^2 = OP^2 - OT^2$
$(iii) \quad PA \cdot PB = PT^2$

Write 'True' or 'False' and give reasons for your answer.
If a chord $AB$ subtends an angle of $60^{\circ}$ at the centre of a circle,then the angle between the tangents at $A$ and $B$ is also $60^{\circ}$.

$A$ tangent $\overline{PM}$ is drawn from a point $P$ outside $\odot(O, 8)$. $\overline{OP}$ intersects the circle at $N$. If $NP = 2$,then $PM = \ldots$

In the figure,common tangents $AB$ and $CD$ to two circles intersect at $E$. Prove that $AB = CD$.

Write 'True' or 'False' and give reasons for your answer.
The length of the tangent from an external point $P$ on a circle with centre $O$ is always less than $OP$.