Prove that the two line<->circle intersecting points multiplied by eachother are equal to the square of the circle's tangent On my current math assignment there is a question as follows:
Prove the following:
A line $k$ through point $P$ intersects a circle in the points $A$ and $B$. Another line $l$ also through $P$ is tangent to the circle in point $C$. Then the lengths $PA$, $PB$ and $PC$ satisfy
$$
PA\cdot PB = PC^2
$$
Can anyone give me any clues as to how to prove this?
 A: Let the center be 'O' and drop a perpendicular from 'O' to AB and let this point be 'D'.
$(PD – AD) (PD + BD)$
$= (PD – AD) (PD + AD) $
$= PD^2 – AD^2$
In right ΔOPD,
$OP^2 = OD^2 + PD^2$
$⇒ PD^2 = OP^2 – OD^2$
∴ $PA . PB = (OP^2 – OD^2) – AD^2 = OP^2 – (OD^2 + AD^2)$
In right ΔOAD,
$OA^2 = OD^2 + AD^2$
∴ $PA . PB = OP^2 – OA^2 = OP^2 – OT^2 (OA = OT)$
In ΔOPT,
$OP^2 = PT^2 + OT^2$
$⇒ OP^2 – OT^2 = PT^2$
∴$ PA . PB = PT^2$
A: 
$PC^2=MP^2-r^2$
$MP^2-r^2 =PA \times AB $ (Why?)
Here's the proof:
$PA=PB-AB$
$AB = 2(r^2 -MX^2)$
$MX^2=MP^2 - (PX^2) \implies MP^2 - PA+ (\dfrac{AB}{2})^2$
Now you can complete the proof yourself.!
A: The requested result is actually the consequence of a theorem called "power of a point".
To prove it, (referring to the drawing), Knowledge required are



*

*angle ABC = angle ACP -- [for the reason of "angles in the alternate segment"].

*$\triangle PCB$ ~ $\triangle PAC$ --[for the reason of "equi-angular"].
Result comes out after setting up the corresponding ratios. 
