Let there is a circle of centre O.

Given AB is a chord of length 8.

Again from A and B, two tangents are drawn and they intersect at point P.

Now join O an P.

Sinec we know that tangents drawn from the external point to a circle are equal in length.

So PA = PB

Hense traingle APB is an isoscale triangle.

Since OP is the bisector of ∠APB, So OP is perpendicular to AB.

Therefore, OP bisects AB.

So AQ = BQ = 8/2 = 4

Now in ΔAOQ,

from Pythagorus Theorem,

       OA² = OQ² + AQ²

=> OQ² = OA² - AQ²

=> OQ² = 5² - 4²     (since OA is radius, So OA = 5)

=> OQ² = 25 - 16

=> OQ² = 9

=> OQ = √9

=> OQ = 3

Agian we know that a tangent is perpendicular to the radius at the point of contact.

So ∠PAQ + ∠OAQ = 90 ...............1

Now in ΔAPQ,

∠APQ + ∠PAQ = 90 .................2

from equation 1 and 2, we get

      ∠PAQ + ∠OAQ = ∠APQ + ∠PAQ

=> ∠OAQ = ∠APQ

So ΔAQP is similar to ΔOQA.

So PA/OA = AQ/OQ

=> PA/5 = 4/3

=> PA = (4*5)/3

=> PA = 20/3

=> PA = 6.7

So value of PA = 6.7 unit