Question

In the given figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q meet at a point T. Find the length of TP.

Answer 1

We know that tangents from an external point are equal in length, therefore TP = TQ

⇒ ΔTPQ is an isosceles triangle.

Also, we know that tangents from the external points are equally inclined to the segment.

∴ OT is bisector of ∠PTQ

⇒ OT ⊥ PQ ....[∵ In an isosceles triangle, angle bisector and altitude are the same]

⇒ ∠ORP = ∠TRP = 90°

Now, In ΔORP,

OR = `sqrt(5^2 - "PR"^2)`

= `sqrt(25 - 4^2)` .....[∵ Perpendicular from centre to a chord bisect the chord and so PR = RQ = 4 cm]

= `sqrt(25 - 16)`

= 3 cm

∵ ∠OPT = 90 .....[∵ Radius is ⊥ to the tangent at the point of contact]

∴ TP² = OT² – OP² = (TR + 3)² – 5²  ...(i)

Also, from ΔPRT,

TP² = PR² + TR² = 4² + TR²  ...(ii)

From equations (i) and (ii), we get

(TR + 3)² – 5² = 4² + TR²

⇒ TR² + 9 + 6TR – TR² = 25 + 16 = 41

⇒ 6TR = 41 – 9 = 32

⇒ TR = `16/3`

Now, from equation (ii), we get

TP² = `4^2 + (16/3)^2`

= `(16 xx 9 + 256)/9`

= `(144 + 256)/9`

= `400/9`

= `20/3`

TP = 6.67 cm