if PAB is a secant to a circle intersecting the circle at A and B and PT is a tangent segment, prove that
PA * PB=PT2
Also, using the above theorem prove the following,
two circles intersect each other at P and Q. from a point R on PQ produced, two tangents RB and RC are drawn to the circles touching hem at B and C
prove tha RC=RB
Hi!
Here is the answer to the question.
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Given: A secant PAB to a circle C (0, r) intersect it in A and B and PT is a tangent segment.
To prove: PA × PB = PT2
Construction: Draw OD ⊥ AB. Join OP, OT and OA
Proof: Since OD ⊥ AB
⇒ AD = DB … (1) (Perpendicular from the centre to the chord bisects the chord)
PA × PB = (PD – AD) (PD + BD)
= (PD – AD) (PD – AD) (from (1))
= PD2 – AD2
In right ΔOPD,
OP2 = OD2 + PD2
⇒ PD2 = OP2 – OD2
∴PA × PB = (OP2 – OD2) – AD2
= OP2 – (OD2 + AD2)
In right ΔOAD
OA2 = OD2 + AD2
∴ PA × PB = OP2 – OA2 = OP2 – OT2 (OA = OT)
In right ∆OPT,
OP2 = PT2 + OT2
⇒ OP2 – OT2 = PT2
∴PA × PB = PT2
Cheers!