(Q2) PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.

PQOT8cm5cm

Given PQ = 8

PR = 4

PO2 = PR2+OR2

= +OR2

OR =

Now let RT = x and PT on triangle OPT ∠p = 90°

OT is hypotenuse

OT2 = OP2+PT2

(Pythagoras theorem)

(+x)2 = 2+y2 --------(1)

and in triangle PRT ∠R= 90°

PT is hypotenuse

PT2 = PR2+RT2

y2 = 2+x2-------(2)

Now putting the value of y2= 42+x2 in equation (1) we got

(+x)2 = 2+x2+2

+x2+6x=++x2

6X = + - = + =

X =
6
 =
3

Now from equation (2) we get

y2 = 2+x2 = + (
3
 )2
= +
9
=
+
9
 =
9
=PT = y = √
9
 =
3
 = .66cm