Answer :
In the given figure, PQ = 16 cm
PO = 10 cm (radius)
⸫ TP = TQ [Tangents drawn from an external point to the same circle are equal]
ΔTPQ is an isosceles triangle
⸫ OT is the perpendicular bisector of PQ
⸫ PR = 8 cm
In ΔTPR,
TP2 = TR2 + PR2 [Pythagoras theorem]
TP2 = TR2 + 82 …. (i)
∠TPO = 90° [A tangent is perpendicular to the radius at the point of tangency]
In ΔTPO,
TO2 = TP2 + PO2 [Pythagoras theorem]
TO2 = TP2 + 102
TP2 = TO2 – 100 …. (ii)
In ΔOPR,
OR2 = OP2 – PR2 [Pythagoras theorem]
OR2 = 102 – 82
⸫ OR = √36 = 6 cm
Equation eq (i) and eq (ii),
TR2 + 64 = TO2 – 100
TR2 + 64 = (TR + OR)2 – 100 [From the figure, (TR + OR) = TO]
TR2 + 64 = TR2 + 36 + 12TR – 100
12TR = 128
⸫ TR = 
Substituting the value of TR in eq (i), we get
TP2 = 
TP2 = 
⸫ TP = 
⸫ TP = 13.33 cm
Rate this question :
Bonus Questions on Circles40 mins
