Answers

2014-03-01T19:55:52+05:30
The center of the circle is O, the tangent is RP and RAB is a secant of the circle.

So, RP2 = RA × RB  .... (1)

Likewise, The second circle with center O', RP' is the tangent like upper one and RAB is a secant to the circle.

So, RP' 2 = RA × RB  .... (2)

Equating both (1) and (2), we will get
RP2 = RP' 2
RP = RP'
Presently, RS2 - AB2 = (RS + AB)(RS - AB)
= (RA + AB + BS + AB)(RA + AB + BS - AB)  [As RS = RA + AB + BS]
= (RA + AB + BS + AB)(RA + AB + BS - AB)
= (RA + BS + 2AB)(RA + BS)  
= (RA + RA + 2AB)(RA + RA)  [As we know from above RA = BS]
= (2 RA + 2 AB)(2 RA)  
= 2(RA + AB)(2 RA)  
= 2(RB)(2 RA)  [by means of: RA + AB = RB]
= 4RA.RB = 4RP2 [using the eq (1)] 
= 4.{(1/2)PP’}^2 = PP' 2
= RS2 - AB2 = PP' 2
= RS2 = PP' 2 + AB2
=RS2 = PP' 2 + AB2
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