Question

PP' and QQ' are 2 direct tangents to two circles intersection at A and B.The common chord is produced which intersect PP' at R and QQ' at S.Prove that i)RA=SB 
                                                                                           ii)RS ^{2} =P'P ^{2}+ AB^{2}

Answer 1

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)] 
$= 4.{(1/2)PP’}^2 = PP' 2$
$= RS2 - AB2 = PP' 2$
$= RS2 = PP' 2 + AB2$
$=RS2 = PP' 2 + AB2$