Chapter 10 · Question 7

$PQ$ is a chord of length $8\text{ cm}$ of a circle of radius $5\text{ cm}$ . The tangents at $P$ and $Q$ intersect at a point $T$ . Find the length $TP$ .

Answer Revealed

Direct Answer:

Join

OT

. Let

OT

meet

PQ

R

. Since

\triangle TPQ

is isosceles and

TO

is the angle bisector of

\angle PTQ

OT \perp PQ

and

PR = RQ = 4\text{ cm}

. In right

\triangle OPR

OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3\text{ cm}

. By AA similarity, right

\triangle TRP \sim \triangle PRO

, giving

\frac{TP}{PO} = \frac{RP}{RO}

, i.e.,

\frac{TP}{5} = \frac{4}{3}

, so

TP = \frac{20}{3}\text{ cm}

Simple Explanation

The length

TP

\frac{20}{3}\text{ cm}

(about

6.67\text{ cm}

). Using the right triangles formed by the centre, the chord, and the tangent intersection point, we can find this length through similarity and Pythagoras.

Exam-Ready Structure

Given: Radius

OP = 5\text{ cm}

, chord

PQ = 8\text{ cm}

. Tangents at

P

and

Q

meet at

T

. To Find:

TP

. Construction: Join

OT

, meeting

PQ

R

. Proof/Calculation: Since

TP = TQ

(Theorem 10.2),

\triangle TPQ

is isosceles.

TO

is the angle bisector of

\angle PTQ

(remark of Theorem 10.2), so

OT

is the perpendicular bisector of

PQ

. Therefore,

OT \perp PQ

and

PR = RQ = \frac{8}{2} = 4\text{ cm}

. In right

\triangle OPR

OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3\text{ cm}

. In right

\triangle TRP

\angle TPR + \angle RPO = 90^\circ

and

\angle TPR + \angle PTR = 90^\circ

(since

\angle TRP = 90^\circ

). So

\angle RPO = \angle PTR

. Thus,

\triangle TRP \sim \triangle PRO

(AA). From similarity:

\frac{TP}{PO} = \frac{RP}{RO}

, i.e.,

\frac{TP}{5} = \frac{4}{3}

. Therefore,

TP = \frac{20}{3}\text{ cm}

. Alternative: Using Pythagoras:

TP^2 = TR^2 + 16

and

TP^2 + 5^2 = (TR + 3)^2

. Subtracting gives

TP = \frac{20}{3}\text{ cm}

Key Points

OT is perpendicular bisector of chord PQ (PR = RQ = 4 cm)
OR = √(5² − 4²) = 3 cm
Similar triangles TRP and PRO: TP/5 = 4/3
TP = 20/3 cm ≈ 6.67 cm

$PQ$ is a chord of length $8\text{ cm}$ of a circle of radius $5\text{ cm}$ . The tangents at $P$ and $Q$ intersect at a point $T$ . Find the length $TP$ .