PP' and QQ' are two direct common tangents to two circles intersecting at points A and B. The common chord on produced intersects PP' on R and QQ' in S. Prove that
(i) RA = SB.
(ii) SR ² = P' P ² + AB ²

Dear student
Q 23

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
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