Circles Class 10 Mathematics — Important Questions

20 Questions | 70 Marks

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 = 13 cm. Find the length of PQ. 5marks

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. 5marks

In Fig. 10.12, PQRS is a quadrilateral circumscribing a circle. Prove that PQ + RS = PS + QR. 3marks

In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°. 5marks

Prove that the tangents drawn at the ends of a diameter of a circle are parallel. 3marks

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. 3marks

The length of a tangent from a point A at distance 10 cm from the centre of the circle is 8 cm. Find the radius of the circle. 2marks

Two concentric circles are of radii 6 cm and 4 cm. Find the length of the chord of the larger circle which touches the smaller circle. 3marks

A parallelogram ABCD is drawn to circumscribe a circle. Prove that the parallelogram is a rhombus. 5marks

A triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 3 cm and 4 cm respectively. Find the lengths of the sides AB and AC. 5marks

In Fig. 10.14, ABC is a triangle circumscribing a circle. If AB = AC, prove that the point of contact D bisects BC. 2marks

From an external point P, tangents PA and PB are drawn to a circle with centre O. If the tangents are inclined at an angle of 60°, find ∠POA. 3marks

A chord of length 16 cm is drawn from a point P to a circle of radius 10 cm. Find the distance of the point P from the centre of the circle. 3marks

From an external point P, two tangents PA and PB are drawn to a circle with centre O. If OP = 10 cm and the radius of the circle is 6 cm, find the length of PA. 3marks

In Fig. 10.11, TP and TQ are two tangents to a circle with centre O. If ∠POQ = 120°, then ∠PTQ is equal to: (A) 60° (B) 70° (C) 80° (D) 90° 1mark

The lengths of tangents drawn from an external point to a circle are: (A) unequal (B) parallel (C) equal (D) perpendicular 1mark

A tangent to a circle intersects it in: (A) no point (B) one point (C) two points (D) infinitely many points 1mark

The line segment joining the centre of a circle to the point of contact is: (A) a tangent (B) a secant (C) a radius (D) a chord 1mark

How many tangents can be drawn to a circle from a point inside the circle? (A) 0 (B) 1 (C) 2 (D) Infinitely many 1mark

If the area of a circle is 154 cm², then the radius of the circle is: (A) 7 cm (B) 14 cm (C) 3.5 cm (D) 10.5 cm 2marks

Answers

We are given a circle with centre O and radius 5 cm. PQ is a tangent at P. OQ = 13 cm. Since the tangent at any point of a circle is perpendicular to the radius through the point of contact, OP ⊥ PQ.In right-angled triangle OPQ, by Pythagoras theorem:

O Q^{2} = O P^{2} + P Q^{2}

13^{2} = 5^{2} + P Q^{2}

169 = 25 + P Q^{2}

P Q^{2} = 169 - 25 = 144

$P Q = \sqrt{144} = 12$ cm. [5 marks]

Let ∠OPQ =

α

. In right-angled triangle OPQ,

∠ O Q P = 90^{\circ} - α

.We know that tangents drawn from an external point to a circle are equal, so TP = TQ.Also, OP = OQ (radii).In

△

TPQ, TP = TQ, so it is an isosceles triangle.In

△

OPQ and

△

OTQ, OQ is common, OP = OT (radii), TQ = TP (tangents). This is incorrect.Consider

△

TPO and

△

TQO.TP = TQ (tangents from an external point)OP = OQ (radii)TO = TO (common side)So,

△

TPO

≅

△

TQO (SSS congruence).Therefore, ∠PTO = ∠QTO and ∠TPO = ∠TQO.In right-angled triangle TPO, ∠TOP + ∠TPO = 90°.Also, in

△

TPQ, let ∠PTQ =

θ

. Since TP = TQ,

△

TPQ is isosceles.∠TPQ = ∠TQP =

\frac{180^{\circ} - θ}{2} = 90^{\circ} - \frac{θ}{2}

.We know that OP ⊥ TP, so ∠OPT = 90°.∠OPQ = ∠OPT - ∠TPQ =

90^{\circ} - (90^{\circ} - \frac{θ}{2}) = \frac{θ}{2}

.So,

θ = 2 \times ∠ O P Q

.Therefore, ∠PTQ = 2∠OPQ. 5marks

Let the points of contact be P, Q, R, S on AB, BC, CD, DA respectively.We know that tangents from an external point to a circle are equal in length.So, AP = AS, BP = BQ, CQ = CR, DR = DS.Now, consider the sum of opposite sides:AB + CD = (AP + PB) + (CR + RD)AD + BC = (AS + SD) + (BQ + QC)Substituting the equal lengths:AB + CD = (AS + BQ) + (CQ + DS)AD + BC = (AS + DS) + (BQ + CQ)Rearranging the terms on the right side:AB + CD = AS + BQ + CQ + DSAD + BC = AS + DS + BQ + CQSince addition is commutative, the sums are equal.Therefore, AB + CD = AD + BC. 3marks

Let the parallel tangents be XY and X'Y'. Let AB be another tangent touching the circle at C. Let AB intersect XY at A and X'Y' at B.Consider

△

OAX and

△

OAC.OA is common.OX = OC (radii).AX = AC (tangents from external point A).So,

△

OAX

≅

△

OAC (SSS).This implies ∠XOA = ∠COA.Similarly, consider

△

OBY and

△

OCB.OB is common.OY = OC (radii).BY = BC (tangents from external point B).So,

△

OBY

≅

△

OCB (SSS).This implies ∠YOB = ∠COB.Since XY is a straight line, ∠XOY = 180°.∠XOY = ∠XOA + ∠AOC + ∠COB + ∠BOY = 180°.Since ∠XOA = ∠COA and ∠YOB = ∠COB, we have:2∠AOC + 2∠COB = 180°.2(∠AOC + ∠COB) = 180°.∠AOC + ∠COB = 90°.∠AOB = 90°. 5marks

Let the diameter be AB. Let P and Q be the points where the diameter meets the circle, so AP and BQ are radii.Let the tangents at A and B be L and M respectively.The radius OA is perpendicular to the tangent L at A. So, OA ⊥ L.The radius OB is perpendicular to the tangent M at B. So, OB ⊥ M.Since AB is a diameter, OA and OB are along the same line. Thus, OA and OB are collinear.L is perpendicular to AB, and M is perpendicular to AB.Lines perpendicular to the same line are parallel.Therefore, L is parallel to M.Hence, the tangents drawn at the ends of a diameter of a circle are parallel. 3marks

Let the tangent to the circle at point P be XY. Let O be the centre of the circle. We need to prove that the perpendicular to XY at P passes through O.By Theorem 10.1, the radius OP is perpendicular to the tangent XY at the point of contact P.So, the line OP is the perpendicular to the tangent XY at P.Thus, the perpendicular at the point of contact to the tangent passes through the centre. 3marks

Let the centre of the circle be O. Let A be the external point and let P be the point of contact of the tangent from A to the circle.We are given OA = 10 cm and AP = 8 cm.Since the radius through the point of contact is perpendicular to the tangent, OP ⊥ AP.In right-angled triangle OPA, by Pythagoras theorem:

O A^{2} = O P^{2} + A P^{2}

10^{2} = O P^{2} + 8^{2}

100 = O P^{2} + 64

O P^{2} = 100 - 64 = 36

$O P = \sqrt{36} = 6$ cm.
The radius of the circle is 6 cm. [2 marks]

Let the two concentric circles have centre O. Let the radii be R = 6 cm and r = 4 cm.Let AB be the chord of the larger circle that touches the smaller circle at point P.Join OP. Since AB is tangent to the smaller circle at P, OP ⊥ AB.Also, AB is a chord of the larger circle. Since OP ⊥ AB, OP bisects AB. So, AP = PB.In right-angled triangle OPA, by Pythagoras theorem:

O A^{2} = O P^{2} + A P^{2}

R^{2} = r^{2} + A P^{2}

6^{2} = 4^{2} + A P^{2}

36 = 16 + A P^{2}

A P^{2} = 36 - 16 = 20

$A P = \sqrt{20} = 2 \sqrt{5}$ cm.
The length of the chord AB = 2 * AP = 2 * $2 \sqrt{5}$ = $4 \sqrt{5}$ cm. [3 marks]

Let ABCD be a parallelogram circumscribing a circle. Let the points of contact on AB, BC, CD, DA be P, Q, R, S respectively.Since tangents from an external point are equal:AP = ASBP = BQCR = CQDR = DSIn a parallelogram, opposite sides are equal: AB = CD and BC = DA.AB = AP + PBCD = CR + RDBC = BQ + QCDA = DS + SASince AB = CD, AP + PB = CR + RD.Since BC = DA, BQ + QC = DS + SA.We have AP = AS, BP = BQ, CR = CQ, DR = DS.Consider AB = CD.AP + PB = CR + RD.Also, BC = DA.BQ + QC = DS + SA.Substitute PB = BQ, QC = CR, RD = DS, SA = AP.BQ + CR = DS + AP.We know AB = CD. So, AP + PB = CR + RD.We know BC = DA. So, BQ + QC = DS + SA.Since ABCD is a parallelogram, AB || CD and BC || DA.Also, AB = CD and BC = DA.Let AP = x, PB = y, CR = z, RD = w.So, AS = x, BQ = y, CQ = z, DS = w.AB = x + yCD = z + wBC = y + zDA = w + xSince AB = CD, x + y = z + w.Since BC = DA, y + z = w + x.From x + y = z + w, we have x - z = w - y.From y + z = w + x, we have z - x = w - y.So, -(x - z) = w - y.This means -(w - y) = w - y, which implies 2(w - y) = 0, so w = y.If w = y, then x + y = z + y, which means x = z.So, we have x = z and y = w.This means AP = AS = CR = CQ = x, and BP = BQ = DR = DS = y.Therefore, AB = x + y, BC = y + x, CD = x + y, DA = y + x.All sides are equal.Hence, the parallelogram is a rhombus. 5marks

Let the triangle be ABC, circumscribing a circle with centre O and radius r = 2 cm. Let the points of contact be D on BC, E on AB, and F on AC.We are given BD = 3 cm and DC = 4 cm.Since tangents from an external point are equal:BE = BD = 3 cmCF = CD = 4 cmLet AE = AF = x cm.The sides of the triangle are:BC = BD + DC = 3 + 4 = 7 cm.AB = AE + EB = x + 3 cm.AC = AF + FC = x + 4 cm.The semi-perimeter s =

\frac{A B + B C + A C}{2} = \frac{(x + 3) + 7 + (x + 4)}{2} = \frac{2 x + 14}{2} = x + 7

cm.The area of

△

ABC can be calculated by Heron's formula:Area =

\sqrt{s (s - a) (s - b) (s - c)}

Area =

\sqrt{(x + 7) (x + 7 - 7) (x + 7 - (x + 4)) (x + 7 - (x + 3))}

Area =

\sqrt{(x + 7) (x) (4) (3)} = \sqrt{12 x (x + 7)}

cm².The area of

△

ABC can also be calculated as Area = r * s, where r is the inradius.Area = 2 * (x + 7) cm².Equating the two expressions for the area:

2 (x + 7) = \sqrt{12 x (x + 7)}

Squaring both sides:

4 (x + 7)^{2} = 12 x (x + 7)

4 (x + 7) (x + 7) = 12 x (x + 7)

Divide by $4 (x + 7)$ (since x+7 cannot be 0):
x + 7 = 3x
2x = 7
x = 3.5 cm.
So, AE = AF = 3.5 cm.
AB = x + 3 = 3.5 + 3 = 6.5 cm.
AC = x + 4 = 3.5 + 4 = 7.5 cm.
The sides are AB = 6.5 cm, AC = 7.5 cm, and BC = 7 cm. [5 marks]

△

ABC, let the inscribed circle touch BC at D, AC at F, and AB at E.We are given AB = AC, so

△

ABC is isosceles.Tangents from an external point to a circle are equal.AE = AFBE = BDCF = CDWe are given AB = AC.AB = AE + EBAC = AF + FCSo, AE + EB = AF + FC.Since AE = AF, we must have EB = FC.We know BE = BD and CF = CD.Therefore, BD = CD.This means that D bisects BC. 2marks

Let P be the external point, and PA and PB be the tangents to the circle with centre O.We are given that the tangents are inclined at 60°, so ∠APB = 60°.In

△

OAP and

△

OBP:OA = OB (radii)OP is common.PA = PB (tangents from an external point).So,

△

OAP

≅

△

OBP (SSS).Therefore, ∠APO = ∠BPO =

\frac{60^{\circ}}{2} = 30^{\circ}

.Also, OP bisects ∠AOB.In right-angled triangle OAP (since radius is perpendicular to tangent at point of contact), ∠OPA = 30°.We need to find ∠POA.In

△

OAP, ∠OAP = 90°.∠POA + ∠OPA + ∠OAP = 180°.∠POA + 30° + 90° = 180°.∠POA + 120° = 180°.∠POA = 180° - 120° = 60°. 3marks

Let the chord be PQ of length 16 cm. Let the point be P. Let the circle have radius 10 cm and centre O.We are looking for the distance of P from the centre, which is OP.The question is phrased confusingly. It seems to imply that a chord of length 16 cm is drawn from a point P, which would mean P is on the circle. However, the context of tangents implies P might be an external point. Let's assume P is an external point and a chord of length 16 cm is drawn such that its endpoints are on the circle, and one of the tangents from P is considered. This is not clear.

Let's interpret the question as: A point P is outside a circle of radius 10 cm. A tangent from P to the circle has length such that a chord related to it is 16 cm. This is still ambiguous.

Re-interpreting based on common question types: "A chord of length 16 cm is drawn in a circle of radius 10 cm. Let P be one of the endpoints of the chord. Find the distance of P from the centre." This is trivial, as P is on the circle, so the distance is the radius, 10 cm.

Let's assume the question meant: "A chord PQ of length 16 cm is drawn in a circle of radius 10 cm with centre O. Find the distance of the chord from the centre."
Let M be the midpoint of PQ. Then OM ⊥ PQ.
In right $△$ OMP, OP (radius) = 10 cm, PM = 16/2 = 8 cm.

O M^{2} + P M^{2} = O P^{2}

O M^{2} + 8^{2} = 10^{2}

O M^{2} + 64 = 100

$O M^{2} = 36$ , so OM = 6 cm.

If the question meant: "From a point P outside a circle of radius 10 cm, a tangent PQ is drawn of length 16 cm. Find the distance of P from the centre."
In right $△$ OPQ, $O P^{2} = O Q^{2} + P Q^{2} = 10^{2} + 16^{2} = 100 + 256 = 356$ .
$O P = \sqrt{356} = 2 \sqrt{89}$ cm.

Given the provided text and typical NCERT questions, it's most likely the question implies a chord of length 16 cm in a circle of radius 10 cm. Let's assume P is an external point and the tangent length is 16 cm.
Let O be the centre, radius r = 10 cm. Let P be the external point. Let PQ be the tangent of length 16 cm.
In right-angled $△$ OPQ, $O P^{2} = O Q^{2} + P Q^{2} = 10^{2} + 16^{2} = 100 + 256 = 356$ .
$O P = \sqrt{356} = 2 \sqrt{89}$ cm. This is likely the intended meaning if "chord of length 16 cm" was a mistake for "tangent of length 16 cm".

Let's re-examine the phrasing: "A chord of length 16 cm is drawn from a point P to a circle of radius 10 cm. Find the distance of the point P from the centre of the circle." This implies P is on the circle, and the distance is the radius (10 cm). This is too simple.

Alternative interpretation: P is an external point. A chord of length 16 cm is related to the tangents from P. This is still too vague.

Let's assume the question meant: "A tangent PQ is drawn from an external point P to a circle of radius 10 cm. If the length of the tangent PQ is 16 cm, find the distance of P from the centre."
In right $△$ OPQ, $O P^{2} = O Q^{2} + P Q^{2} = 10^{2} + 16^{2} = 100 + 256 = 356$ .
$O P = \sqrt{356} = 2 \sqrt{89}$ cm. [3 marks]

Let O be the centre of the circle and P be the external point. Let PA be the tangent.We are given OP = 10 cm and the radius OA = 6 cm.Since the radius to the point of contact is perpendicular to the tangent, OA ⊥ PA.In right-angled triangle OAP, by Pythagoras theorem:

O P^{2} = O A^{2} + P A^{2}

10^{2} = 6^{2} + P A^{2}

100 = 36 + P A^{2}

P A^{2} = 100 - 36 = 64

$P A = \sqrt{64} = 8$ cm. [3 marks]

Let O be the centre of the circle and T be the external point. Let TP and TQ be the tangents.We are given ∠POQ = 120°.Consider the quadrilateral TPQO. The sum of angles is 360°.∠TPQ + ∠PQO + ∠QOT + ∠TPO = 360° (Incorrect statement of angles in a general quadrilateral)We know that the radius to the point of contact is perpendicular to the tangent.So, ∠TPO = 90° and ∠TQO = 90°.In quadrilateral TPQO, the sum of angles is:∠PTQ + ∠TPQ + ∠PQO + ∠QOT = 360° (Incorrect combination of angles)Correct angles in quadrilateral TPQO are:∠PTQ + ∠TPO + ∠POQ + ∠OQT = 360°∠PTQ + 90° + 120° + 90° = 360°∠PTQ + 300° = 360°∠PTQ = 360° - 300° = 60°.The correct option is (A). 1mark

Tangents drawn from an external point to a circle are equal in length. 1mark

A tangent to a circle intersects it at exactly one point. 1mark

The line segment joining the centre of a circle to the point of contact is the radius. 1mark

From a point inside a circle, no tangent can be drawn. Any line passing through an interior point will be a secant, intersecting the circle at two points. 1mark

The area of a circle is given by

A = π r^{2}

.We are given the area

A = 154

cm².Using

π = \frac{22}{7}

\frac{22}{7} r^{2} = 154

r^{2} = \frac{154 \times 7}{22}

r^{2} = 7 \times 7 = 49

$r = \sqrt{49} = 7$ cm.
The radius of the circle is 7 cm. [2 marks]

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Circles — Important Questions

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