Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 10 Circles Important Questions and Answers.

## RBSE Class 10 Maths Chapter 10 Important Questions Circles

Objective Type Questions—

Question 1.

The length of a tangent from an external point A to a circle of radius 6 cm is 10 cm. The distance of the centre of the circle from point A will be—

(A) 7 cm

(B) 8 cm

(C) 9 cm

(D) 10 cm

Answer:

(B) 8 cm

Question 2.

A point P is at a distance of 25 cm from the centre of a circle. The radius of the circle is 7 cm. Then the length of the tangent drawn from P to the circle will be—

(A) \(\sqrt {30}\) cm

(B) 24 cm

(C) 28 cm

(D) 30 cm

Answer:

(B) 24 cm

Question 3.

If all the four sides of a rhombus ABCD touch a circle then—

(A) AC + DA = BD + CD

(B) AB + CD = BC + DA

(C) AB + CD = AC + BC

(D) AC + AD = BC + BD

Answer:

(B) AB + CD = BC + DA

Question 4.

A line which cuts a circle in two points is called—

(A) chord

(B) Tangent

(C) Secant

(D) None of these

Answer:

(C) Secant

Question 5.

The maximum number of tangents that can be drawn from an external point to a circle is—

(A) One

(B) Two

(C) Three

(D) Infinitely many

Answer:

(B) Two

Question 6.

The tangents drawn at the ends of a diameter of a circle are—

(A) Perpendicular

(B) Parallel

(C) Intersecting

(D) None of these

Answer:

(B) Parallel

Question 7.

In the given figure AB and AC are tangents to the circle, O is the centre of the circle, if ∠CAB = 60°, then ∠BDC is—

(A) 60°

(B) 70°

(C) 120°

(D) 150°

Answer:

(A) 60°

Very Short Answer Type Questions—

Question 1.

In the given figure the circumcircle of △ABC touches the sides at Q, P and R. If AQ = 8cm, then find the perimeter of △ABC.

Solution:

We know that

AQ = \(\frac{1}{2}\) Perimeter (△ABC)

⇒ Perimeter (AABC) = 2 × AQ

= 2 × 8 = 16 cm.

Question 2.

In the given figure find the value of QS when it is given that OQ = 13 cm. and PQ = 12 cm.

Solution:

According to the figure PQ is a tangent.

∴ ∠OPQ = 90°

So now in △OPQ

(OP)^{2} + (PQ)^{2} = (OQ)^{2}

⇒ (OP)^{2} + (12)^{2} = (13)^{2}

⇒ (OP)^{2} = 169 – 144 = 25

∴ OP = \(\sqrt {25}\) = 5 cm.

Question 3.

In figure if TP and TQ are two tangents to circle with centre O such that ∠POQ = 150° then find the value of ∠PTQ.

Solution:

PT and TQ are tangents.

∴ ∠POQ + ∠PTQ = 180°

⇒ 130° + ∠PTQ =180°

⇒ ∠PTQ = 180° – 130° = 50°

Question 4.

In the given figure, from an external point A two tangents AB and AC are drawn to a circle. If ∠BAC = 48°, then write the value of ∠ABC.

Solution:

∠B + ∠C= 180° – 48°

= 132°

But ∠B = ∠C

∵ Tangents are equal in length.

∴ ∠ABC = \(\frac{132^{\circ}}{2}\)

= 66°

Question 5.

In the figure, AB, BC and CA are tangents to a circle. If BC = 6.3 cm and MC = 2.7 cm, then write the measure of BL.

Answer:

We know that

MC = NC

[The lengths of the tangents are equal.]

∴ BN = (6.3 – 2.7)

BN = 3.6 cm.

and BN = BL

∴ BL = 3.6 cm.

Question 6.

Two circles touch externally of the radii of the two circles are 5 cm and 3 cm, then write the distance between their centres.

Answer:

5 + 3 = 8 cm.

Question 7.

Two circles touch each other internally, then write the number of their common tangents.

Answer:

One.

Question 8.

In the given figure PA and PB are tangents to a circle. ∠APB = 40°, then write thd value of ∠AOB.

OR

It tangents RA and RB from a point R to a circle with centre O are inclined to each other at an angle of θ and ∠AOB = 40° then find the value of θ.

Answer:

∠AOB = 180° – 40° = 140°

Question 9.

In the given figure PBQ is tangent to a circle at point B. Write the value of ∠ABP.

Answer:

∠ABP = 90°

Question 10.

The length of a tangent drawn from any external point to a circle is 12 cm, if the radius of the circle is 5 pm, then find the distance of the external point from the centre.

Answer:

Distance = \(\sqrt{(12)^{2}+(5)^{2}}\)

= \(\sqrt {169}\)

= 13 cm.

Question 11.

In the given figure O is the centre of a circle. AB and AC are tangents to the circle. If OA = 10 cm and OB = 6 cm then write the length of AC.

Solution:

AB = \(\sqrt{10^{2}-6^{2}}\)

= \(\sqrt{100-36}\)

= \(\sqrt{64}\) = 8 cm

and AB AC So, AC = 8 cm.

Question 12.

In the figure TP and TQ are tangents to a circle with centre O. If ∠TOQ = 50°, then find ∠OTP.

Solution:

∵ QT is a tangent

∴ ∠OQT = 90°

From △OQT∠OTQ

= 180° – (∠OQT+ ∠QOT)

= 180° – (50° + 90°)

= 180° – 140° = 40°

∵ △QOT = △POT (∵ PT = QT)

∠P = ∠Q = 90°

or = OT

∴ ∠OTP = ∠OTQ

So, .∠OTP = 40°

Question 13.

How many tangents can be drawn at a point lying on the circumference of a circle.

OR

How many tangents can be constructed to any point on the circle of radius 4 cm?

Answer:

One and only one tangent can be drawn at any point lying on the circmference of a

Question 14.

In the given figure, if PA and PB are two tangents to a circle with centre O such that angle APB = 80°, then find the value of angle AOB.

Solution:

In the figure PA and PB are two tangents to a circle with centre O.

∴ ∠PAO = ∠PBO = 90°

and ∠APB = 80° [Given]

∴ ∠AOB = 360° – (90° + 90° + 80°)

= 360° – 260° = 100°

Question 15.

If from a point A, two tangents AB and AC to a circle with centre O are such that ∠BOC = 140°, then write the value of ∠BAC.

Solution:

In the figure, AB and AC are two tangents to a circle with centre O.

∴ ∠ABO = ∠ACO = 90°

and ∠BOC = 140° [Given]

∴ ∠BAC + ∠BOC = 180°

So ∠BAC = 360° – (90° + 90° + 140°)

= 360° – 320° = 40°

Question 16.

The tangent PQ at a point P of a circle with radius 1.5 cm meets a line. Through the centre O at point Q such that OQ = 13 cm, then find the length of PQ.

Solution:

Drawing figure according to the question

OP = 5 cm. and OQ 13 cm.

∵ PQ is a tangent and OP is a radius.

∴ ∠OPQ = 90°

Now in right △OPQ,

By Pythagoras Theorem,

(OQ)^{2} = (OP)^{2} + (PQ)^{2}

Putting the value,

(13)^{2} = (5)^{2} + (PQ)^{2}

or 169 = 25 + (PQ)^{2}

or (PQ)^{2} = 169 – 25 = 144

or PQ = \(\sqrt {144}\) = 12 cm

Question 17.

If the tangents TA and TB drawn from a point T to a circle with centre O are inclined at an angle of 70° with each other, then find ∠AOB.

Solution:

According to the figure TA and TB are two tangents from any point T to a circle with centre O.

∴ ∠TAO = ∠TBO = 90°

and ∠ATB – 70°

∴ ∠AOB = 360° – (90° + 90° + 70°)

= 360° – 250°

= 110°

Short Answer Type Questions—

Question 1.

Prove that the length of two tangents drawn to a circle from any external point are equal.

OR

Prove that the lengths of the tangents drawn from an external point to a circle are equal.

In the given figure, O is the centre of a circle and two tangents KR, KS are drawn on the circle from a point K lying outside the circle. Prove that KR = KS.

Solution:

Given—The centre of the circle is O and two tangents from an external point R are RP and RQ.

To Prove RP = RQ

Construction : Join OP, OQ and OR

Proof : We know that a tangent line is perpendicular to the radius through the point of contact.

So, ∠OPR = ∠OQR = 90° …. (i)

Now in △OPR and △OQR,

∠OPR = ∠OQR | From equation (i)

OR = OR (Common side)

OP = OQ (radii of the same circle)

∴ By right angle – hypotenuse – side congruence property

△OPR ≅ △OQR

Therefore the corresponding sides of congruent triangles will be equal.

⇒ RP = RQ (Hence Proved)

Question 2.

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

Solution:

Given : We are given two concentric circles C_{1} and C_{2} with centre O and a chord AB of the larger circle C_{1} which touches the smaller circle C_{2} at the point P.

To Prove AP = BP

Construction : Join OP

Proof : AB is a tangent to C_{2} at P and OP is its radius. Therefore by theorem,

OP ⊥ AB

Now, AB is a chord of the circle C_{1} and OP ⊥ AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre to the chord bisects the chord,

i.e., AP = BP (Hence Proved)

Question 3.

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that

∠PTQ = 2∠OPQ

Solution:

Given—We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P and Q are the points of contact, (see figure)

To Prove—∠PTQ = 2∠OPQ

Proof—∠PTQ = θ

Now by Theorem

TP = TQ

So, TPQ is an isoscles triangle.

So, ∠TPQ = ∠TQP = \(\frac{1}{2}\)(180° – θ)

= 90° – \(\frac{1}{2}\)θ

By Theorem, ∠OPT = 90°

So, ∠OPQ = ∠OPT – ∠TPQ

= 90°- (90° – \(\frac{1}{2}\)θ)

= \(\frac{1}{2}\)θ = \(\frac{1}{2}\) ∠PTQ

i.e., ∠PTQ = 2 ∠OPQ (Hence Proved)

Question 4.

If two tangent lines be drawn to a circle from an external point then prove that those lines subtend equal angles at the centre of the circle and also prove that both the tangent lines make equal angles with the line segment joining the centre of the circle to the external point.

Solution:

Given—A circle C (0, r) or which two tangent lines AP and AQ have been drawn from an external point A.

To Prove—(i) ∠AOP = ∠AOQ

(ii) ∠OAP = ∠OAQ

Construction—Join OA, OP and OQ

Proof—In △OPA and △OQA

AP = AQ

(Tangents to the same circle)

OA= OA (Common side)

OP = OQ (Radii)

By S-S-S Congruence

△OPA = △OQA

⇒ ∠AOP = ∠AOQ

and ∠OAP = ∠OAQ (Hence Proved)

Question 5.

Prove that the tangent lines drawn at the end points of any chord of a circle make equal angles with the chord.

Solution:

Let the tangents PA and PB be drawn at the end points of a chord AB of circle C(0, r) which intersect at point P.

Let OP intersect chord AB at point C.

To Prove— ∠PAC = ∠PBC

Proof—In △PCA and △PCB

PA = PB

| (Tangents lines from an external point P to the circle)

PC = PC (Common side)

∠APC = ∠BPC

| [∵ Tangent line PA and PB make equal angles with OP]

By S-A-S Congruence

△PCA ≅ △PCB

⇒ ∠PAC ≅ ∠PBC (CPCT) (Hence Proved)

Question 6.

Two tangent lines AB and AC are drawn to a circle with centre O from an external point O. Prove that ∠BAC = 2∠OBC.

Solution:

We are given a circle with centre O, an external point A and two tangent lines AB and AC to the circle, where B, C are the points of contact as shown in the figure.

To prove- ∠BAC = 2∠OBC

Let ∠BAC = θ

By Theorem we know that

AB = AC

So, ABC is an isosceles triangle

∴ ∠ABC = ∠ACB = \(\frac{1}{2}\)(180° – θ)

∠ABC = 90° – \(\frac{1}{2}\)θ

∠OBA = 90°

So ∠OBC = ∠OBA – ∠ABC

= 90° – (90° – \(\frac{1}{2}\)θ)

= 90° – 90° + \(\frac{1}{2}\)θ

∠OBC = \(\frac{1}{2}\)∠BAC

∴ ∠BAC = 2∠OBC Hence Proved

Long Answer Type Questions—

Question 1.

In the figure O is the centre of the circle and the tangent lines are PA and PB, then prove that AOBP is a cyclic quadrilateral.

Solution:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Now OA ⊥ AP and OB ⊥ BP

⇒ ∠OAP = 90° and ∠OBP = 90°

⇒ ∠OAP + ∠OBP = 90° + 90°

⇒ ∠OAP + ∠OBP = 180° …. (i)

The sum of the angles of the quadrilateral OAPB is 360°.

∴ ∠APB + ∠BOA = 360° – (180°)

∠APB + ∠BOA = 180° …. (ii)

From equation (i) and (ii) it is clear that

∠OAP + ∠OBP = 180°

and ∠APB + ∠BOA – 180°

Here the sum of the opposite angles is 180°. Hence the quadrilateral OABP is a cyclic quadrilateral. (Hence Proved)

Question 2.

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. (see figure). Find the length TP.

Solution:

Join OT. Let it intersect PQ at the point R. Then ATPQ is isosceles and TO is the angle bisector of ∠PTQuestion So, OT ⊥ PQ and therefore, OT bisects PQ which gives

PR = RQ = 4 cm.

Also, OR = \(\sqrt{\mathrm{OP}^{2}-\mathrm{PR}^{2}}\)

= \(\sqrt{5^{2}-4^{2}}\) cm.

3 cm.

We can determine TP by Pythagoras Theorem as follows :

Let TP = x and TR = y then

In right △PRT,

(TP)^{2} = (TR)^{2} + (PR)^{2}

(x)^{2} = y^{2} + (4)^{2}

x^{2} = y^{2} + 16 ……(i)

In right △OPT

(OT)^{2} = (PT)^{2} + (PO)^{2}

⇒ (TR + OR)^{2} = (x)^{2} + (5)^{2}

⇒ (y + 3)^{2} = x^{2} + 25 …. (ii)

Subtracting equation (i) fr om equation (ii)

x^{2} + 25 – x^{2} = (y + 3)^{2} – y^{2} – 16

⇒ 25 = y^{2} + 6y + 9 – y^{2} – 16

⇒ 25 = 6y – 7

⇒ 6y = 32

y = \(\frac{32}{6}\) = \(\frac{16}{3}\)

From equation (i)

Question 3.

A line intersecting two tangent lines PA and B from any external point P to a circle touches the circle at point M, then prove that

PL + LM = PN + NM

OR

In figure PA and PB are tangent lines of a circle. If M be a point on the circle, then prove that

PL + LM = PN + NM

Solution:

Given—A circle where centre is OP is a point and side the circle, PA and PB are two tangent lines. M is any point on the circle. Line segment LN passes through the point M.

To Prove— PL + LM = PN + NM

Proof : Two tangent lines PA and PB are drawn from P to a circle.

∴ PA = PB

or PL + LA = PN + NB …. (i)

LA and LM are two tangent lines to the circle from point L

∴ LA = LM …. (ii)

Similarly, NB = NM …. (iii)

From (i), (ii) and (iii)

PL + LM = PN + NM (Hence Proved)

Question 4.

The circle drawn with side AB of a right angled triangle ABC as diameter intersects the hypotenuse AC at P. Prove that the tangent to the circle at point P intersects the side BC.

Solution:

To Prove : BQ = QC,

Where Q is the point of intersection of tangent drawn at P and BC.

Construction—Join BP

Proof— AB is a diameter of the circle.

∴ ∠APB = 90°

(Angle in a semi-circle is a right angle)

∴ ∠APB + ∠BPC = 180°

∴ 90° + ∠BPC = 180°

∴ ∠BPC = 90° …. (i)

∵ In △ABC ∠ABC = 90°

∴ ∠BAC + ∠ACB = 90° …. (ii) (∵ The sum of the three angles of a triangle is 180°.)

From equations (i) and (ii)

∠BPC = ∠BAC + ∠ACB

or ∠BPQ + ∠CPQ = ∠BAC + ∠ACB …. (iii)

But ∠BPQ = ∠BAC

(Angle in the alternate segment)

∴ From equation (iii)

∠CPQ= ∠ACB

⇒ ∠CPQ = ∠PCQ

(∵ ∠ACB = ∠PCQ)

∴ PQ=QC …. (iv)

Again PQ= QB …. (v)

(Tangent lines drawn from an external point to a circle are equal.)

∴ From equations (iv) and (v)

QC= QB

Hence, PQ intersects BC.

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