Practice Questions: Circles | Mathematics (Maths) Class 10 PDF Download

Sol:

From the diagram:-

RA + AO = 12     ______ (1)

RC + CP = 13 _______ (2)          

OB + BP = 5 _______ (3)

Now from the above figure  

Side AO = Side OB = Radius of the circle

Side RA = Side RC  

Line BP = Line CP

Rewriting these three equations we get

RA + AO = 12 _______ (4)

RA + BP = 13 _______ (5)

AO + BP = 5 _______(6)

(4) – (6) + (5)

RA – BP = 7

RA + BP = 13

2 RA = 20

RA = 10

From equation 4, AO = 12 – 10 = 2 cm

Since AO is equal to radius of the inscribed circle, radius = 2 cm.

Sol: Two tangents can be drawn from the external point to a circle.

Sol: It is given that PA and PB are tangents to the given circle.

∴ ∠PAO = 90° (Radius is perpendicular to the tangent at the point of contact.)

Now, ∠PAB = 50° (Given)
∴ ∠OAB = ∠PAO − ∠PAB = 90° − 50° = 40°

In △OAB,

OB = OA (Radii of the circle)
∴ ∠OAB = ∠OBA = 40° (Angles opposite to equal sides are equal.)

Now
∠AOB + ∠OAB + ∠OBA = 180° (Angle sum property)
⇒ ∠AOB = 180° − 40° − 40° = 100°

Sol: In the given figure,
In △ACO,

OA = OC …(Radii of the same circle)

∴ △ACO is an isosceles triangle.

∠CAB = 30° …(Given)

∴ ∠CAO = ∠ACO = 30°…(angles opposite to equal sides of an isosceles triangle are equal)

∠PCO = 90° …(radius drawn at the point of contact is perpendicular to the tangent)

Now ∠PCA = ∠PCO − ∠CAO

∴ ∠PCA = 90° − 30° = 60°

Sol: Given that AB = 12 cm, BC = 8 cm and AC = 10 cm.

Let AD = AF = p cm, BD = BE = q cm and CE = CF = r cm ....(Tangents drawn from an external point to the circle are equal in length)

⇒ 2(p + q + r) = AB + BC + AC = AD + DB + BE + EC + AF + FC = 30 cm

⇒ p + q + r = 15

AB = AD + DB = p + q = 12 cm

Therefore, r = CF = 15 - 12 = 3 cm.

AC = AF + FC = p + r = 10 cm

Therefore, q = BE = 15 - 10 = 5 cm.

Therefore, p = AD = p + q + r - r - q = 15 - 3 - 5 = 7 cm.

Sol: From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is 60 cm².

Explanation:

Firstly, draw a circle of radius 5 cm with centre O.

P is a point at a distance of 13 cm from O.

A pair of tangents, PQ and PR, is drawn.

Thus, quadrilateral PQOR is formed.

∴ OQ ⊥ QP ...[Since QP is a tangent line]

In right-angled ΔPQO,

OP² = OQ² + QP²

⇒ 13² = 5² + QP²

⇒ QP² = 169 - 25 = 144

⇒ QP = 12 cm

Now, the area of ΔOQP

= ½ × QP × QO

= ½ × 12 × 5

= 30 cm²

∴ Area of quadrilateral PQOR

= 2 × ar ΔOQP

∴ Area of quadrilateral PQOR

= 2 × ar ΔOQP

= 2 × 30

= 60 cm²

Sol: False 

Explanation:

​As tangent to any point on the circle is perpendicular to the radius through the point of contact,

OA ⏊ AC and OB ⏊ CB

∠OBC = ∠OAC = 90° …(1)

Using angle sum property of quadrilateral in Quadrilateral AOBC,

∠OBC + ∠OAC + ∠AOB + ∠ACB = 360°

Substituting the values

90° + 90° + 60° + ∠ACB = 360°

So we get

∠ACB = 120°

Angle between two tangents is 120°.

Therefore, the statement is false.

Sol: False 
Explanation:

Let P be a point external to the circle away from O

Construct a tangent PA on the circle

PA > r

Now, let us consider a point P1 on the tangent AP near to contact point A of tangent

PA, P1A < AO

The length of tangent PA and P1A are greater and smaller than the radius OA

So the length of tangent may or may not be greater than the radius

Therefore, the statement is false.

Sol: (i) A tangent to the circle cuts it into one point(s).

(ii) A line intersecting the circle in two points is called a secant.

(iii) A circle could have two parallel tangents at the most.

(iv) The common point of the tangent to the circle and the circle is called the point of contact.

Sol: We have to find the length of the tangent PQ.

ΔOPQ is a right-angle triangle according to Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

By Pythagoras theorem:

OQ² = OP² + PQ²

12² = 5² + PQ²

144 = 25 + PQ²

PQ² = 119

PQ = √119

Thus, the length of PQ is √119 cm.

Sol: Let's draw a figure as per the question.

The lengths of tangents drawn from an external point to a circle are equal.

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

In ΔOAP and in ΔOBP

OA = OB (radii of the circle are always equal)

AP = BP (length of the tangents)

OP = OP (common)

Therefore, by SSS congruency ΔOAP ≅ ΔOBP

SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

If two triangles are congruent then their corresponding parts are equal.

Hence,

∠POA = ∠POB

∠OPA = ∠OPB

Therefore, OP is the angle bisector of ∠APB and ∠AOB

Hence, ∠OPA = ∠OPB = 1/2 (∠APB )

= 1/2 × 80°

= 40°

By angle sum property of a triangle,

In ΔOAP

∠A + ∠POA + ∠OPA = 180°

OA ⊥ AP (Theorem 10.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.)

Therefore, ∠A = 90°

90° + ∠POA + 40° = 180°

130° + ∠POA = 180°

∠POA = 180° - 130°

∠POA = 50°

Sol: The chord of the larger circle is a tangent to the smaller circle as shown in the figure below.

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle

PQ is a chord of a larger circle and a tangent of a smaller circle.

Tangent PQ is perpendicular to the radius at the point of contact S.

Therefore, ∠OSP = 90°

In ΔOSP (Right-angled triangle)

By the Pythagoras Theorem,

OP² = OS² + SP²

5² = 3² + SP²

SP² = 25 - 9

SP² = 16

SP = ± 4

SP is the length of the tangent and cannot be negative

Hence, SP = 4 cm.

QS = SP (Perpendicular from center bisects the chord considering QP to be the larger circle's chord)

Therefore, QS = SP = 4cm

Length of the chord PQ = QS + SP = 4 + 4

PQ = 8 cm

Therefore, the length of the chord of the larger circle is 8 cm.

Sol: Given, PA and PB are the two tangents drawn to a circle with centre O though an external point P.

At point E on the circle the tangent is drawn which intersects PA and PB at C and D.

Given, PA = 10 cm

We have to find the perimeter of the triangle PCD.

We know that the tangents drawn to a circle through an external point are equal.

From the figure,

The tangents drawn through point P are PA and PB

So, PA = PB -------------- (1)

The tangents to the circle through C are CA and CE

So, CA = CE -------------- (2)

The tangents to the circle through D are DB = DE

So, DB = DE -------------- (3)

Considering triangle PCD,

Perimeter = PC + PD + CD

From the figure,

CD = CE + DE

So, perimeter = PC + PD + CE + DE

From (2) and (3),

Perimeter = PC + PD + CA + DB

On rearranging,

Perimeter = PC + CA + PD + DB

From the figure,

PC + CA = PA

PD + DB = PB

By (1), PA = PB

So, perimeter = PA + PB

Perimeter = PA + PA

= 2PA

Given, PA = 10 cm

So, 2PA = 2(10)

= 20 cm

Therefore, the perimeter of the triangle PCD is 20 cm.

Sol: False

Given, PQL and PRM are the tangents to the circle with centre O at the points Q and R

S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°

We have to determine if ∠QSR is equal to 40°.

We know that the radius of circle is perpendicular to the tangent at the point of contact.

So, ∠LQO = ∠MRO = 90°

From the figure,

∠LQO = ∠LQS + ∠SQO

90° = 50° + ∠SQO

∠SQO = 90° - 50°

∠SQO = 40°

From the figure,

∠MRO = ∠MRS + ∠SRO

90° = 60° + ∠SRO

∠SRO = 90° - 60°

∠SRO = 30°

Considering triangle OSQ,

OS = OQ = radius

Two sides of a triangle are equal

So, OSQ is an isosceles triangle.

In an isosceles triangle, two sides and two angles are equal.

Two equal angles are ∠SOQ = ∠ORS

∠QSR = ∠OSQ + ∠OSR

∠QSR = 30° + 40°

Therefore, ∠QSR = 70°

Sol: Given: PA and PB are the tangents to the circle.

PA = 12 cm

QC = QD = 3 cm

To find: PC + PD

PA = PB = 12 cm (The lengths of tangents drawn from an external point to a circle are equal)

Similarly, QC = AC = 3 cm and QD = BD = 3 cm.

Now, PC = PA - AC = 12 - 3 = 9 cm

Similarly, PD = PB - BD = 12 - 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.