PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of the TP.
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Given: 

Length of PQ = 4.8 cm

Radius of the circle = 3 cm

To find: Here we need to find the length of the TP.

Solution:

The lengths of tangents drawn from an external point to a circle are equal. So,

Then △TPQ is isosceles and TO is the angle bisector of ∠PTQ and in Isosceles triangle angle bisector is altitude also.

Therefore,       ∠PRT = ∠PRO = 90^o   

So,

PR = RQ    (Perpendicular from center to a chord  bisect the chord)

PR = RQ = $\frac{1}{2}$PQ =  $\frac{1}{2}$ $\times$ 4.8 = 2.4 cm

Using Pythagoras theorem in △PRO:

OR² = OP² $-$ PR²   

OR² = 3² $-$ 2.4²   

OR² = 9 $-$ 5.76

OR = $\sqrt{3.24}$

OR = 1.8 cm

In △TPR if ∠PRT = 90^o  

Then,         

∠TPR $+$ ∠PTR = 90^o      …..(1)

Also,

∠TPO = 90^o          (The tangent at any point of a circle is perpendicular to the radius through the point of contact)

∠TPR $+$ ∠RPO = 90^o         …..(2) 

From (1) and (2):

∠TPR $+$ ∠PTR = ∠TPR $+$ ∠RPO

∠PTR = ∠RPO       …..(3)

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Updated on: 10-Oct-2022

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