PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangents at P and Q intersect at a point T as shown in the figure.
Find the length of TP.
Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T
To Find : Length of PT
Construction : Join OQ
Now in △OPT and △OQT
OP = OQ
[radii of same circle]
PT = PQ
[tangents drawn from an external point to a circle are equal]
OT = OT
[Common]
△OPT ≅ △OQT
[By Side - Side - Side Criterion]
∠POT = ∠OQT
[Corresponding parts of congruent triangles are congruent]
or ∠POR = ∠OQR
Now in △OPR and △OQR
OP = OQ
[radii of same circle]
OR = OR [Common]
∠POR = ∠OQR [Proved Above]
△OPR ≅ △OQT
[By Side - Angle - Side Criterion]
∠ORP = ∠ORQ
[Corresponding parts of congruent triangles are congruent]
Now,
∠ORP + ∠ORQ = 180°
[Linear Pair]
∠ORP + ∠ORP = 180°
∠ORP = 90°
⇒ OR ⏊ PQ
⇒ RT ⏊ PQ
As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have
PR = QR = PQ/2 = 16/2 = 8 cm
∴ In right - angled △OPR,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
(OP)2 = (OR)2 + (PR)2
(10)2 = (OR)2 + (8)2
100 = (OR)2 + 64
(OR)2= 36
OR = 6 cm
Now,
In right angled △TPR, By Pythagoras Theorem
(PT)2 = (PR)2 + (TR)2 [1]
Also, OP ⏊ OT
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In right angled △OPT, By Pythagoras Theorem
(PT)2 + (OP)2 = (OT)2
(PR)2 + (TR)2 + (OP)2= (TR + OR)2 [From 1]
(8)2 + (TR)2 + (10)2 = (TR + 6)2
64 + (TR)2 + 100 = (TR)2 + 2(6)TR + (6)2
164 = 12TR + 36
12TR = 128
TR = 10.7 cm [Appx]
Using this in [1]
PT2 = (8)2 + (10.7)2
PT2 = 64 + 114.49
PT2 = 178.49
PT = 13.67 cm [Appx]