PQ is a chord of lenght 8cm of a circle of radius 5cm and centre O. The tangents at P and Q intersects at point T. Find length of TP
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Join OTOT
In \Delta OPTΔOPT
OT^2=OP^2+PT^2\cdots(1)OT
2
=OP
2
+PT
2
⋯(1)
OTOT is perpendicular bisector to PQPQ let they met at SS
\implies SP=4⟹SP=4cm
By pythagorean theorm
\implies SO=3⟹SO=3cm
By property associated with altitudes of Right angled triangle
\implies \Delta OPT\approx\Delta PSO⟹ΔOPT≈ΔPSO
\implies \dfrac{OS}{SP}=\dfrac{OP}{PT}\implies \dfrac{3}{4}=\dfrac{5}{TP}\implies TP=6.66⟹
SP
OS
=
PT
OP
⟹
4
3
=
TP
5
⟹TP=6.66 cm
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