Tangent Properties of Circles — Analytical & Application Based Questions

In the given figure O is the centre of the circle. ABCD is a quadrilateral whose sides AB, BC, CD and DA touch the circle at E, F, G and H respectively. If AB = 15 cm, BC = 18 cm and AD = 24 cm, find the length of CD.

Answer

We know that,

The length of tangents drawn from an external point to the circle are equal.

⇒ AE = AH = x (let)

⇒ BE = BF = y (let)

⇒ CF = CG = z (let)

⇒ DG = DH = k (let)

Given,

⇒ AB = 15

⇒ AE + EB = 15

⇒ x + y = 15 .......(1)

⇒ BC = 18

⇒ BF + CF = 18

⇒ y + z = 18 ........(2)

⇒ AD = 24

⇒ AH + DH = 24

⇒ x + k = 24 ..........(3)

Adding equations (2) and (3), we get :

⇒ y + z + x + k = 18 + 24

⇒ x + y + z + k = 42

⇒ 15 + z + k = 42 [From equation (1)]

⇒ z + k = 42 - 15

⇒ z + k = 27.

⇒ CD = CG + DG = k + z = 27 cm.

Hence, length of CD = 27 cm.

In the given diagram, ABCDEF is a regular hexagon inscribed in a circle with centre O. PQ is a tangent to the circle at D. Find the value of :

(a) ∠FAG

(b) ∠BCD

(c) ∠PDE

Answer

(a) By formula,

Each interior angle of a n sided polygon = $\dfrac{(n - 2) \times 180°}{n}$

$= \dfrac{(6 - 2) \times 180°}{6} \\[1em] = \dfrac{4 \times 180°}{6} \\[1em] = \dfrac{2}{3} \times 180° \\[1em] = 120°.$

∴ ∠FAB = 120°

From figure,

∠FAG and ∠FAB form a linear pair.

∴ ∠FAB + ∠FAG = 180°

⇒ 120° + ∠FAG = 180°

⇒ ∠FAG = 180° - 120° = 60°.

Hence, ∠FAG = 60°.

(b) Each interior angle of regular hexagon = 120°.

Hence, ∠BCD = 120°.

(c) In triangle DEF,

⇒ ∠DEF = 120° (Each interior angle of a regular hexagon equals to 120°)

⇒ DE = EF (Sides of a regular hexagon)

⇒ ∠EDF = ∠EFD = a (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠DEF + ∠EDF + ∠EFD = 180°

⇒ 120° + a + a = 180°

⇒ 2a = 180° - 120°

⇒ 2a = 60°

⇒ a = $\dfrac{60}{2}$ = 30°

⇒ ∠EFD = 30°.

By alternate segment theorem,

The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle subtended by that chord in the alternate segment of the circle.

∴ ∠PDE = ∠EFD = 30°.

Hence, ∠PDE = 30°.

In the adjoining diagram PQ, PR and ST are the tangents to the circle with centre O and radius 7 cm. Given OP = 25 cm. Find :

(a) length of ST

(b) value of ∠OPQ, i.e. θ

(c) ∠QUR, in nearest degree

Answer

(a) We know that,

The radius of a circle and tangent are perpendicular at the point of contact.

∴ ∠PQO = 90°

In right-angled triangle PQO,

⇒ PO² = PQ² + OQ²

⇒ 25² = PQ² + 7²

⇒ 625 = PQ² + 49

⇒ PQ² = 625 - 49

⇒ PQ² = 576

⇒ PQ = $\sqrt{576}$ = 24 cm.

In △ POQ,

⇒ tan θ = $\dfrac{OQ}{PQ}$

⇒ tan θ = $\dfrac{7}{24}$ ...............(1)

From figure,

⇒ ∠PAS = 90°

⇒ PA = OP - OA = 25 - 7 = 18 cm.

⇒ tan θ = $\dfrac{AS}{PA}$

⇒ tan θ = $\dfrac{AS}{18}$ ...........(2)

From equation (1) and (2), we get :

$\Rightarrow \dfrac{7}{24} = \dfrac{AS}{18} \\[1em] \Rightarrow AS = \dfrac{7}{24} \times 18 \\[1em] \Rightarrow AS = \dfrac{7 \times 3}{4} = \dfrac{21}{4} \text{ cm}.$

From figure,

⇒ ST = 2 × AS = $2 \times \dfrac{21}{4} = \dfrac{21}{2}$ = 10.5 cm

Hence, ST = 10.5 cm.

(b) From equation (1),

⇒ tan θ = $\dfrac{7}{24}$

⇒ tan θ = 0.292

⇒ tan θ = tan 16° 16'

⇒ θ = 16° 16'.

Hence, θ = 16° 16'.

(c) In △ POQ,

⇒ ∠POQ + ∠PQO + ∠QPO = 180°

⇒ ∠POQ + 90° + 16° 16' = 180°

⇒ ∠POQ + 106° 16' = 180°

⇒ ∠POQ = 180° - 106° 16' = 73° 44' = 74°.

We know that,

Tangent from an external point to the circle are equal in length.

∴ PQ = PR.

Also,

OR = OQ (Both equal to radius of circle)

In △ POQ and △ POR,

⇒ PQ = PR (Proved above)

⇒ OQ = OR (Radius of same circle)

⇒ PO = PO (Common side)

∴ △ POQ ≅ △ POR (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠POR = ∠POQ = 74°

From figure,

⇒ ∠QOR = ∠POR + ∠POQ = 74° + 74° = 148°.

We know that,

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.

⇒ ∠QOR = 2∠QUR

⇒ ∠QUR = $\dfrac{∠QOR}{2} = \dfrac{148°}{2}$ = 74°.

Hence, ∠QUR = 74°.

In the given figure, angle ABC = 70° and angle ACB = 50°. Given, O is the centre of the circle and PT is the tangent to the circle. Then calculate the following angles

(a) ∠CBT

(b) ∠BAT

(c) ∠PBT

(d) ∠APT

Answer

Join AT and BT.

(a) We know that,

Angle in a semicircle is a right angle.

∴ ∠CBT = 90°.

Hence, ∠CBT = 90°.

(b) In cyclic quadrilateral ATBC,

⇒ ∠CBT + ∠CAT = 180° (∵ Sum of opposite angles of a cyclic quadrilateral = 180°)

⇒ 90° + ∠CAT = 180°

⇒ ∠CAT = 180° - 90° = 90°.

In △ABC,

⇒ ∠CBA + ∠CAB + ∠ACB = 180° [By angle sum property of triangle]

⇒ 70° + ∠CAB + 50° = 180°

⇒ ∠CAB + 120° = 180°

⇒ ∠CAB = 180° - 120°

⇒ ∠CAB = 60°.

From figure,

∠BAT = ∠CAT - ∠CAB = 90° - 60° = 30°.

Hence, ∠BAT = 30°.

(c) From figure,

∠BTX = ∠BAT = 30° [Angle in same segment are equal]

∠PBT = ∠CBT - ∠CBA = 90° - 70° = 20°.

Hence, ∠PBT = 20°.

(d) Since, ∠PTB and ∠BTX are linear pairs.

⇒ ∠PTB = 180° - ∠BTX = 180° - 30° = 150°.

In △PBT,

⇒ ∠PBT + ∠PTB + ∠APT = 180° [By angle sum property of triangle]

⇒ 20° + 150° + ∠APT = 180°

⇒ ∠APT + 170° = 180°

⇒ ∠APT = 180° - 170°

⇒ ∠APT = 10°.

Hence, ∠APT = 10°.