Description

This mock test of Test: Circles- 4 for UPSC helps you for every UPSC entrance exam.
This contains 25 Multiple Choice Questions for UPSC Test: Circles- 4 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Circles- 4 quiz give you a good mix of easy questions and tough questions. UPSC
students definitely take this Test: Circles- 4 exercise for a better result in the exam. You can find other Test: Circles- 4 extra questions,
long questions & short questions for UPSC on EduRev as well by searching above.

QUESTION: 1

The length of tangent PQ, from an external point P is 24 cm. If the distance of the point P from the centre is 25 cm, then the diameter of the circle is

Solution:

Here ∠OPQ = 90°[Angle between tangent and radius through the point of contact]

∴ OQ^{2} = OP^{2} + PQ^{2} (25)^{2} = OP^{2} + (24)^{2} ⇒ OP^{2} = 625 - 576 ⇒ OP^{2} = 49

⇒ OP = 7 cm

Therefore, the diameter = 2 x OP = 2 x 7 = 14 cm

QUESTION: 2

A tangent PQ at point of contact P to a circle of radius 12 cm meets the line through centre O to a point Q such that OQ = 20 cm, length of tangent PQ is :

Solution:

Since op is perpendicular to PQ, the ∠OPQ = 90°

Now, in right angled triangle OPQ,

QUESTION: 3

PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such ∠POR = 120^{∘}, then ∠OPQ is

Solution:

QUESTION: 4

A circle is inscribed in ΔABC having sides 8 cm, 10 cm and 12 cm as shown in the figure. Then,

Solution:

Let AD = x and BE = y

∴ BD = 12 - x ⇒ BE = 12 - x [BD = BE = Tangents to a circle from an external point]

⇒ y = 12 - x ⇒ x+y = 12.......(i)

Also, AF = x and CF = 10 - x and CE = 8 - y

Also, AF = x and CF = 10 — x and CE = 8 - y

∴ 10 - x = 8 - yx - y = 2 (ii)

On solving eq. (i) and (ii), we get x = 7 and y = 5

⇒ AD = 7 cm and BE = 5cm

QUESTION: 5

In the given figure, If TP and TQ are two tangents to a circle with centre O, so that ∠POQ = 110^{o }then ∠PTQ is equal to :

Solution:

Since the angle between the two tangents drawn from an external point to a circle in supplementary of the angle between the radii of the circle through the point of contact.

∴ ∠PTQ = 180^{∘}−110^{∘} = 70^{∘}

QUESTION: 6

A tangent PQ at a point P o a circle of radius 5 cm meets a line through the centre O at point Q, so that OQ = 12 cm. find the length PQ.

Solution:

∠OPQ = 90° [Angle between tangent and radius through the point of contact]

QUESTION: 7

From a point Q, the length of tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm, radius of circle is :

Solution:

Here ∠OPQ = 90° [Tangent makes right angle with the radius at the point of contact]

∴ OQ^{2} = OP^{2} + PQ^{2} (25)^{2} = OP^{2} + (24)^{2}

⇒ OP2 = 676 - 576 = 49

⇒ OP = 7 cm Therefore, the radius of the circle is 7 cm

QUESTION: 8

If PA and PB are tangents to the circle with centre O such that ∠APB = 40^{∘}, then ∠OAB is equal to

Solution:

Let ∠OAB = ∠OBA = x [Opposite angles of opposite equal radii] And ∠AOB =180° - 40° = 140°

Now, in triangle AOB,

∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x +140° = 180°

⇒ 2x = 40°

⇒ x = 20°

∴ ∠OAB = 20°

QUESTION: 9

Quadrilateral PQRS circumscribes a circle as shown in the figure. The side of the quadrilateral which is equal to PD + QB is

Solution:

PD + QB = PA + QA [Tangents from an external point to a circle are equal]

⇒PD + QB = PQ

QUESTION: 10

In the given figure, the pair of tangents A to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm, then the radius of the circle is :

Solution:

Construction: Joined OP and OQ.

Here OP ⊥ AP and OC⊥AQ and PA ⊥ AQ Also AP = AQ

Therefore. APOQ is a square. ⇒ AP = OP = OQ = 5 cm

QUESTION: 11

In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and RQ = 4 cm, then OR is equal to

Solution:

Here SQ = 6 cm, then OQ = OS = 6/2 = 3 cm [Radii] and ∠Q = 90°

Now, in triangle OQR,OR^{2} = QR^{2} + OQ^{2}

⇒ OR^{2} = (4)^{2} + (3)^{2} = 16 + 9 = 25

⇒ OR = 5cm

QUESTION: 12

In figure, AB is a chord of a circle and AT is a tangent at A such that ∠BAT = 60^{o}, measure of ∠ACB is :

Solution:

Since OA is perpendicular to AT. then ∠OAT = 90°

⇒ ∠OAB + ∠BAT = 90°

⇒ ∠OAB + 60° = 90° ⇒ ∠OAB = 30°

∴ ∠OAB = ∠O8A = 30" [Angles opposite to radii]

∴ ∠OAB = 180° - (30° +30° ) = 120° [Angle sum property of a triangle]

∴ Reflex ∠AOB = 360° - 120° = 240°

Now, since the degree measure of an arc of a circle is twice the angle subtended by it any point of the alternate segment of the circle with respect to the arc.

∴ Reflex ∠AOB = 2∠ACB ⇒ 240° = 2∠ACB ⇒ ∠ACB = 120°

QUESTION: 13

If two tangents inclined at 60^{∘} are drawn to circle of radius 3 cm, then length of each tangent is equal to

Solution:

Let O be the centre. Construction: Joined OP.

Since OP bisects ∠P, therefore, ∠APO = ∠OPB = 30° and ∠OAP = 90°

cm Since each tangent from an external point to a circle are equal. Therefore, PA = PB = 3√3 cm

QUESTION: 14

The perimeter of ΔPQR in the given figure is

Solution:

Since Tangents from an external point to a circle are equal.

∴ PA = PB = 4 cm.

BR = CR = 5 cm CQ = AQ = 6 cm ⇒ Perimeter of ΔPQR = PQ + QR + RP

⇒ Perimeter of ΔPQR = PA + AQ + QC + CR + BR + PB ⇒ Perimeter of ΔPQR = 4 + 6 + 6 + 5 + 5 + 4

⇒ Perimeter of Δ PQR = 30 cm

QUESTION: 15

In the given fig., if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50^{∘} with PQ, then ∠POQ is equal to :

Solution:

Since Op is perpendicular to PR, then ∠OPR = 90°

∠RPQ + ∠QPO = 90°

⇒ 50° + ∠QPO = 90°

∠QPO = 40°

Now, OP = OQ (Radii of same circle]

∴ ∠OPQ = ∠OQP = 40°[Angles opposite to equal sides]

In triangle OPQ,

∠POQ + ∠OPQ + ∠OQP = 180°

∠POQ + 40° + 40° = 180°

= ∠ POQ = 100°

QUESTION: 16

In the adjoining figure, the measure Of PR is

Solution:

Here ∠Q = 90° [Angle between tangent and radius through the point of contact] Now, in triangle OPQ, OP^{2} = QO^{2} + PQ^{2}

⇒ OP^{2} = (6)^{2} + (8)^{2} = 36 + 64 = 100

⇒ OP = 10 cm

∴ PR = OP + OR = OP + OQ [OR = OQ = Radii]

⇒ PR = 10+6 = 16cm

QUESTION: 17

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80^{o}, ∠POA is

Solution:

QUESTION: 18

If PQR is a tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 60^{∘}, then ∠AQB is equal to

Solution:

Since AB II PA and BQ is intersectin.

∴ ∠BQR = LQBA = 60° [Alternate angles]

And ∠BQR = ∠QAB = 60° [Alternate segment theorem] Now, in triangle AQB,

∠AQB + ∠QBA + ∠BAQ = 180°

⇒ ∠AQ,B + 60° + 60° = 180°

⇒ ∠LAQB = 60°

QUESTION: 19

In the given figure, a circle touches all four sides of a quadrilateral PQRS, whose sides are PQ = 6.5 cm, QR = 7.3 cm, and PS = 4.2 cm, then RS is equal to

Solution:

Let point of contact of RS be A, point of contact of QR be B, point of contact of PQ be C and point of contact of PS be D. Also let AS = x and AR = y

Now, AS = SD = x [Tangents from an external point]

PD = 4.2 - x

But PD = PC= 4.2 - x

And QC = 6.5 - PC = 6.5 - 4.2 + x = 2.3 + x .........(i)

Now, again, AR = BR = y [Tangents from an external point]

⇒ QB = 7.3 - y

But QB = QC [Tangents from an external point]

QUESTION: 20

In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30^{o}. Then AT is equal to :

Solution:

Construction : Joined OA.

Since OA is perpendicular to AT. then ∠OAP = 90°

In right angled triangle OAT

QUESTION: 21

In the figure given alongside the length of PR is

Solution:

Here ∠Q = 90° and LS = 90° [Angle between tangent and radius through the point of contact]

Now, in triangle OPQ, OP^{2} = OQ^{2} + QP^{2}

⇒ OP^{2} = (3)^{2} + (4)^{2}

⇒ OP^{2} = 16 + 9 = 25

⇒ OP = 5 cm Again in triangle RSO', O'R^{2} = O'S^{2} + FS^{2}

⇒ O'R^{2} = (5)^{2} + (12)^{2}

⇒ O'R^{2} = 25 +144 = 169

⇒ O'R = 13 cm

∴ PR = OP+OQ+O's+OR = 5+ 3+ 5+ 13 = 26 m

QUESTION: 22

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q, such that OQ = 15 cm. Length of PQ is

Solution:

Since OP is perpendicular to PQ, then ∠OPQ = 90°

Now, in triangle OPQ.

OQ^{2} = OP^{2} + PQ^{2} ⇒ (15)^{2} = (5)^{2} + PQ^{2} ⇒ PQ^{2} = 225 - 25

⇒ PQ^{2} = 200 ⇒ PQ = 10√2cm

QUESTION: 23

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80^{∘}, then find ∠POA.

Solution:

Here ∠OAP = 90°

And ∠OPA = 1/2 ∠BPA [Centre lies on the bisector of the angle between the two tangents]

Now, in triangle OPA,

∠OAP + ∠OPA + ∠POA = 180°

⇒ 90° + 40° + ∠POA =180°

⇒ ∠POA = 50°

QUESTION: 24

In the given figure, perimeter of quadrilateral ABCD is

Solution:

Here SD = RD = 5 units [Tangents from an external point]

And PB = QB = 4 cm [Tangents from an external point]

∴ QC = 10 - QB = 10 - 4 = 6 units

Also QC = RC = 6 units [Tangents from an extemal point]

∴ CD = RD + RC = 5 + 6 = 11 units

Also AP = AS = 2 units [Tangents from an external point]

∴ AD = AS + DS = 2 + 5 = 7 units Therefore. Perimeter of quadrilateral ABCD = 6 + 10 + 11 + 7 = 34 units

QUESTION: 25

At one end of a diameter PQ of a circle of radius 5 cm, tangent XPYis drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is

Solution:

Here, OP = 00 = 5 cm [Radii]

And OR = PR - OP = 8 - 5 = 3 cm Also, OA = 5 cm [Radius]

Now, in triangle AOQ, OA^{2} = OR^{2} + AR^{2} ⇒ 52 = 32 + AR^{2}

⇒AR^{2} = 25-9= 160 AR= 4cm

Since, perpendicular from centre of a circle to a chord bisects the chord.

∴ AB = AR+ BR = 4+ 4 = 8cm

### Circles Geometry (Part - 4)

Video | 12:18 min

### Evolution Test-4

Doc | 1 Page

### Solution- Matrices Test- 4

Doc | 5 Pages

### Example: Area Related to Circles- 4

Video | 10:00 min

- Test: Circles- 4
Test | 25 questions | 25 min

- Test: Area Related To Circles- 4
Test | 25 questions | 25 min

- Test: Circles- 2
Test | 20 questions | 40 min

- Test: Area Of Circles
Test | 14 questions | 15 min

- Test: Circles- 2
Test | 25 questions | 25 min