$\odot(O, 41)$ and $\odot(O, 9)$ are concentric circles. The chord $\overline{AB}$ of $\odot(O, 41)$ touches $\odot(O, 9)$ at point $M$. Then $AB = \ldots$

Write 'True' or 'False' and give reasons for your answer.
$AB$ is a diameter of a circle and $AC$ is its chord such that $\angle BAC = 30^{\circ}$. If the tangent at $C$ intersects $AB$ extended at $D$,then $BC = BD$.

$\overline{PA}$ is a tangent to $\odot(O, 8)$ drawn from a point $P$ outside the circle. If $m\angle AOP = 45^\circ$,then $AP = \ldots$

$A$ tangent to a circle forms an angle of measure $\ldots \ldots \ldots \ldots$ with the radius drawn at the point of contact. (in $^{\circ}$)

The tangents drawn from a point $P$ outside $\odot(O, 5)$ touch the circle at $A$ and $B$. If $PA = 8$,then $PB = \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$. $A$ circle touches all the sides of $\Delta ABC$. If $AB + AC = 32$ and $BC = 24$,then find the radius of the incircle.