"PP' and QQ'are two direct common tangents to circle to two circles with centre O and O'intersecting at A ad B .The common chord AB on producing meets the tangents PP'at R and QQ' at S .Show that RS 2 =PP' 2 +AB 2 . "
Given, PP' and QQ' are two direct common tangents to circle to two circles with center O and O' intersecting at A ad B respectively.
To prove: RS2 = PP' 2 + AB2
Proof:
In a circle with center O, RP is a tangent and RAB is a secant to the circle.
Therefore, RP2 = RA × RB .... (1)
Similarly, in the circle with center O', RP' is a tangent and RAB is a secant to the circle.
⇒ RP' 2 = RA × RB .... (2)
On equating (1) and (2), we get
RP2 = RP' 2
⇒ RP = RP'
Now, 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 RA = BS]
= (2 RA + 2 AB)(2 RA)
= 2(RA + AB)(2 RA)
= 2(RB)(2 RA) [using: RA + AB = RB]
= 4RA.RB
= 4RP2 [using (1)]
= 4. = PP' 2
⇒ RS2 - AB2 = PP' 2
⇒ RS2 = PP' 2 + AB2