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

Direct Answer

Join

OT

. Let

OT

meet

PQ

at

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

is

\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

at

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 $TP$ .

Answer

Direct Answer

Simple Explanation

Exam-Ready Structure

Key Points

Related Questions

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$ . Prove that $\angle PTQ = 2\angle OPQ$ .

PQPQPQ is a chord of length 8 cm8\text{ cm}8 cm of a circle of radius 5 cm5\text{ cm}5 cm. The tangents at PPP and QQQ intersect at a point TTT. Find the length TPTPTP.

Answer

Direct Answer

Simple Explanation

Exam-Ready Structure

Key Points

Related Questions

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.

Two tangents TPTPTP and TQTQTQ are drawn to a circle with centre OOO from an external point TTT. Prove that ∠PTQ=2∠OPQ\angle PTQ = 2\angle OPQ∠PTQ=2∠OPQ.

$PQ$ 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 $TP$ .

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$ . Prove that $\angle PTQ = 2\angle OPQ$ .