# CRE 2 - Tangents & Secants | Geometry - Circles

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**CRE 2 - Tangents & Secants | Geometry - Circles**

Two chords AB and CD of a circle intersect at a point P, which lies outside the circle. If PA = 6 cm, PB = 12 cm and PC = 3 cm, what is the length of CD in cm?

- (a)
21

- (b)
24

- (c)
6

- (d)
12

Answer: Option A

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**Explanation** :

For a circle, if two chords AB and CD intersect at a single point P outside the circle, then,

PA × PB = PC × PD

∴ 6 × 12 = 3 × PD

∴ PD = 24 cm

∴ CD = PD – PC = 24 – 3 = 21 cm.

Hence, option (a).

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the given figure, PBQ is a straight line while PA, PB, QB, and QC are tangents to the circle with center O. Find the value of angle AOC.

- (a)
70°

- (b)
80°

- (c)
60°

- (d)
100°

- (e)
None of these

Answer: Option E

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**Explanation** :

In ∆PAB,

PA = PB …(tangents from an external point to a circle are congruent)

∴ m ∠PBA = m ∠PAB; …(angles opposite to congruent sides)

∴ m ∠PBA = 60° …(sum of angles in a triangle is 180°)

Similarly, in ∆QBC

m ∠CBQ = m ∠BCQ = 50°

∵ m ∠PBA + m ∠CBQ + x = 180° …(linear angles)

∴ x = 180° – 60° – 50° = 70°

Now, m ∠AOC = 2 × x

∴ m ∠AOC = 140°

Hence, option (e).

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the figure, PA and PC are tangents to the circle ABC. If ∠P = 48°, then ∠ABC = ?

Answer: 114

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**Explanation** :

Draw two lines joining A and C with center O.

Now in quadrilateral AOCP, sum of all four angles should be 360°.

Also, ∠OAP = ∠OCP = 90° (radius is line joining center and point of tangency is perpendicular to the tangent)

∴ 90° + ∠AOC + 90° + 48° = 360°

⇒ ∠AOC = 132°

Angle subtended by chord AC in the major segment is half of the angle it subtends at the center.

∴ ∠ADC = ½ × ∠AOC = 66°

Now, angles subtended by a chord in different segments are supplementary.

∴ ∠ADC + ∠ABC = 180°

⇒ ∠ABC = 180° - 66° = 114°.

Hence, 114.

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the following figure, lines AP, AQ and BC are tangent to the circle. The length of AP = 22. Find the perimeter of triangle ABC.

Answer: 44

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**Explanation** :

We know 2 tangents drawn from an external point to a circle are equal in length.

⇒ AP = AQ = 22

∴ If we consider point B as the external point, we have two tangents i.e., BP and BD.

⇒ BP = BD

Similarly, if we consider point C as the external point, we have two tangents i.e., CD and CQ.

⇒ CD = CQ

Now, Perimeter of ∆ABC = AB + BD + CD + AC

= AB + BP + CQ + AC

= AP + AQ

= 22 + 22 = 44

Hence, 44.

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the following figure, if PC = 6, CD = 9, PA = 5 and AB = x, find the value of x?

Answer: 13

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**Explanation** :

When two secants are drawn from an external point to a circle

⇒ PC × PD = PA × PB

∴ 6 × 15 = 5 × (5 + AB)

⇒ AB = 13

Hence, 13.

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the following figure, PC = 9, PB = 12, PA = 18, and PF = 8. Then, find the length of DE.

Answer: 2.5

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**Explanation** :

If we consider the smaller circle, two secants are drawn from P.

∴ PC × PB = PF × PE

⇒ 9 × 12 = 8 × PE

⇒ PE = 27/2

Now, if we consider the bigger circle, two secants are drawn from P.

∴ PB × PA = PE × PD

⇒ 12 × 18 = 27/2 × PD

⇒ PD = 16

∴ DE = PD – PE = 16 – 27/2 = 16 – 13.5 = 2.5

Hence, 2.5.

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the figure given below (not drawn to scale), A, B and C are three points on a circle with center O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ∠ATC = 30° and ∠ACT = 50°, then ∠BOA is (in degrees)?

Answer: 100

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**Explanation** :

Given, ∠ATC = 30° and ∠ACT = 50°

In ∆ACT, external angle ,

∠CAB = ∠ACT + ∠ATC = 50° + 30° = 80° …(1)

Also, chord CA and tangent CT make an angle of 50°, hence ∠CBA = 50°.

[Angle made by a chord and tangent is same as the angle subtended by the chord in opposite segment.]

Now, in ∆ABC, sum of all angles = 180°

⇒ ∠CBA + ∠BAC + ∠ACB = 180°

⇒ 50° + 30° + ∠ACB = 180°

⇒ ∠ACB = 100°

Now, angle make by a chord in major segment is half the angle it subtends at the center.

∴ ∠AOB = 2 × ∠ACB

⇒ ∠AOB 2 × 100° = 200°

Hence, 100.

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the following figure, if PT = 6 and AB = 5, find PB?

Answer: 9

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**Explanation** :

Let PA = x

When a secant and a tangent are drawn from an external point to a circle

⇒ PA × PB = PT^{2}

∴ x × (x + 5) = 6^{2}

⇒ x^{2} + 5x – 36 = 0

⇒ (x + 9)(x - 4) = 0

∴ x = -9 or 4. (x = -9 is rejected)

∴ PB = x + 5 = 9.

Hence, 9.

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**CRE 2 - Tangents & Secants | Geometry - Circles**

Two circles of radius 4 and 6 cm and centers P and Q respectively, touch each other externally. From the center of first circle a tangent PR is drawn to second circle which touches the second circle at R. Find PR (in cm)?

[Type in your answer as the nearest possible integer]

Answer: 8

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**Explanation** :

We can draw the following figure from the given information.

In ∆PQR,∠PRQ = 90° (tangent makes an angle of 90° with the radius.)

∴ ∆PQR is a right triangle with PQ as the hypotenuse

⇒ PQ^{2} = PR^{2} + QR^{2}

⇒ 10^{2} = PR^{2} + 6^{2}

⇒ PR^{2} = 100 - 36 = 64

⇒ PR = 8

Hence, 8.

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**CRE 2 - Tangents & Secants | Geometry - Circles**

PQ is the chord of a circle whose center is O. ROS is a line segment originating from a point R on the circle that intersect PQ produced at point S such that QS = OR. If ∠QSR = 30°, then what is the value (in degrees) of POR?

- (a)
30

- (b)
45

- (c)
60

- (d)
90

Answer: Option D

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**Explanation** :

We can draw the following figure from the given information.

Since OR = OP = OQ = OS (radius)

⇒ ∠OPQ = ∠OQP & ∠QSO = ∠QOS

⇒ ∠POQ = 60°

⇒ ∠POR = 180 - (60° + 30°) = 90°

Hence, option (d).

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**CRE 2 - Tangents & Secants | Geometry - Circles**

In the figure given below, O is the center of the circle. OQ is perpendicular to RS and ∠SRT = 30°. If RS = 10√2, then what is the value of PR^{2}?

- (a)
200(1 + √3)

- (b)
300(2 + √3)

- (c)
200(2 + √3)

- (d)
100(3 + 2√3)

Answer: Option C

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**Explanation** :

RQ = QS = 1/2 × 10√2 = 5√2

∴ ∠ORQ = 90 - 30 = 60°

∆ORQ is a 30°-60°-90° triangle

∴ Hypotenuse will be twice the side opposite to 30°

⇒ OR = 10√2 = radius

⇒ OQ = √3/2 × OR = 5√6

In ∆PRQ,

PR^{2} = PQ^{2} + RQ^{2}

⇒ PR^{2} = (10√2 + 5√6)^{2} + (5√2)^{2}

⇒ PR^{2} = 200 + 150 + 100√12 + 50

⇒ PR^{2} = 400 + 200√3 = 200(2 + √3)

Hence, option (c).

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