"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:** RS^{2} = PP' ^{2} + AB^{2}

**Proof:**

In a circle with center O, RP is a tangent and RAB is a secant to the circle.

Therefore, RP^{2} = 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

RP^{2} = RP' ^{2}

⇒ RP = RP'

Now, RS^{2} - AB^{2} = (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

= 4RP^{2} [using (1)]

= 4. = PP' ^{2}

⇒ RS^{2} - AB^{2} = PP' ^{2}

⇒ RS^{2} = PP' ^{2 }+ AB^{2}

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