Circles, Chords and Tangents MCQ Quiz - Objective Question with Answer for Circles, Chords and Tangents - Download Free PDF

Given:

The sum of their areas =  74 πsq cm

Distance between their centers = 12 cm.

Formula Used:

Area of circle = πr²

Calculation:

Let assume that radius of circle 1 = x

So, radius of circle 2 = 12 - x

Area of circle 1 = π(x)²

Area of circle 2 = π(12 - x)²

According to question ⇒ π(x)² + π(12 - x)² = 74π

⇒ x² + 144 - 24x + x² = 74 

⇒ 2x² - 24x + 70 = 0

⇒ x² - 12x + 35 = 0

⇒ (x - 7)(x - 5) = 0

⇒ x = 7 ⇒ x = 5 

∴ The radius of smaller circle is 5 cm

Given:

PQ is the diameter of a circle with center 'O'. Tangent drawn at P meets QR (produced) at S. ∠PSQ = 48º.

Formula Used:

Angle in the semicircle = 90º

Calculation:

In triangle PQR:

Since PQ is diameter ⇒ ∠PRQ = 90°

In triangle PSQ:

∠PSQ is the exterior angle to triangle PQR

∠PSQ = 48º

∠SPQ = 90° (angle made by tangent and chord)

∠PQS = 90 - 48 = 42° 

∠PQS = ∠PQR = 42° 

∴ The value of ∠PQR is 42°.

Given:

PQ is a chord in the minor segment of a circle.

R is a point on the minor arc PQ.

Tangents at points P and Q meet at point T.

∠PRQ = 102°

Formula used:

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

The sum of opposite angles in a cyclic quadrilateral is 180°.

The angle between the tangent and the chord through the point of contact is equal to the angle in the alternate segment (Tangent-Chord Theorem).

The sum of angles in a quadrilateral is 360°.

Calculation:

Let O be the center of the circle.

Consider the cyclic quadrilateral formed by P, R, Q and another point S on the major arc PQ.

The angle subtended by the chord PQ at any point in the major segment will be supplementary to ∠PRQ.

The angle subtended by the major arc PQ at the circumference is ∠PRQ = 102°.

The angle subtended by the minor arc PQ at the circumference (on the major segment) would be 180° - 102° = 78°.

∠PSQ = 180° - ∠PRQ (as PRQS is a cyclic quadrilateral if S is on the major arc).

So, ∠PSQ = 180° - 102° = 78°.

The angle subtended by the minor arc PQ at the center O, i.e., ∠POQ, is twice the angle subtended by it at the circumference in the major segment (∠PSQ).

⇒ ∠POQ = 2 × ∠PSQ

⇒ ∠POQ = 2 × 78°

⇒ ∠POQ = 156°

Now, consider the quadrilateral TP OQ. TP and TQ are tangents to the circle at P and Q, respectively.

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

The sum of angles in the quadrilateral TP OQ is 360°.

⇒ ∠PTQ + ∠TPO + ∠POQ + ∠TQO = 360°

⇒ ∠PTQ + 90° + 156° + 90° = 360°

⇒ ∠PTQ + 336° = 360°

⇒ ∠PTQ = 360° - 336°

⇒ ∠PTQ = 24°

∴ The correct answer is option 2.

Given:

AB || CD (AB is parallel to CD)

Angle between the transversal and AB = 80°

Angle between the other transversal and CD = 20°

Angle at O = x

Concept Used:

When two parallel lines are intersected by a transversal, the alternate interior angles are equal.

The sum of the angles in a triangle is 180°.

Construction:

Draw a line EF parallel to AB and CD passing through point O.

Calculation:

Since AB || EF, the alternate interior angles are equal.

Therefore, ∠BAO = ∠AOE = 80°

Since CD || EF, the alternate interior angles are equal.

Therefore, ∠DCO = ∠COF = 20°

Now, consider the angle at point O, which is x (∠AOC). This angle is the sum of ∠AOE and ∠COF.

x = ∠AOE + ∠COF

x = 80° + 20°

x = 100°

∴ The value of x is 100°.

Given:-

(1) AT = 6 cm, AB = 10 cm & TB = ?

(2) TB ┴ AT

In right-angled, △ATB

Formula used:-

By Pythagoras theorem

(AB)² = (AT)² + (TB)²   

Calculation:-

⇒ (10)² = 6² + (TB)²

⇒ TB = 8

The radius (r) of the circle is TB = 8 cm

We know diameter is the largest chord of the circle

Given∶

AB ∥ CD, and

AB = 10cm, CD = 24 cm

Radii OA and OC = 13 cm

Formula  Used∶

Perpendicular from the centre to the chord, bisects the chord.

Pythagoras theorem.

Calculation∶

Draw OP perpendicular on AB and CD, and 

AB ∥ CD, So, the points O, Q, P are collinear.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

AP = 1/2 AB = 1/2 × 10 = 5cm

CQ = 1/2 CD = 1/2 × 24 = 12 cm

Join OA and OC

Then, OA = OC = 13 cm

From the right ΔOPA, we have

OP² = OA² -  AP²      [Pythagoras theorem]

⇒ OP² = 13²- 5²

⇒ OP2 = 169 - 25 = 144

⇒ OP = 12cm

From the right ΔOQC, we have

OQ² = OC²- CQ²      [Pythagoras theorem]

⇒ OQ2 = 13² - 12²

⇒ OQ2 = 169 - 144 = 25

⇒ OQ = 5 

So, PQ = OP - OQ = 12 -5 = 7 cm

∴ The distance between the chord is of 7 cm.

Concept:

Radius is perpendicular to the tangent at the point of contact

Sum of all the angles of a Quadrilateral = 360° 

Calculation:

PA and PB are tangents drawn from an external point P to the circle.

∠OAP = ∠OBP = 90°  (Radius is perpendicular to the tangent at the point of contact)

Now, In quadrilateral OAPB,

∠APB + ∠OAP + ∠AOB + ∠OBP = 360° 

75° + 90° + ∠AOB + 90° = 360°

∠AOB = 105°

Thus, the angle between the two radii, OA and OB is 105°

Given:

Radius of each circle = 7 cm

BD = transverse common tangent between two circles = 48 cm

Concept used:

Length of direct transverse tangents = √(Square of the distance between the circle - Square of sum radius of the circles)

Length of the direct common tangents =√(Square of the distance between the circle - Square of the difference between the radius of circles)

Calculation:

AC = Length of the direct common tangents

BD = Length of direct transverse tangents

Let, the distance between two circles = x cm

So, BD = √[x² - (7 + 7)²]

⇒ 48 = √(x2 - 142)

⇒ 48² = x2 - 196  [Squaring on both sides]

⇒ 2304 = x2 - 196

⇒ x2 = 2304 + 196 = 2500

⇒ x = √2500 = 50 cm

Also, AC = √[502 - (7 - 7)2]

⇒ AC = √(2500 - 0) = √2500 = 50 cm

∴ The length of BD is 48 cm, length of AC is 50 cm

We know,

Length of direct common tangent = √[d² - (R - r)²]

where d is the distance between the centers and R and r are the radii of the circles.

PQ = √[(R + r)² - (R - r)²]

⇒ PQ = √[R² + r² + 2Rr - (R² + r² - 2Rr)]

⇒ PQ = √4Rr

⇒ PQ² = 4Rr

Concept:

Secant Property

If chord AB and chord CD of a circle intersect at a point P, then

PA × PB = PC × PD

Calculation:

Let the radius of the circle = r 

PA = 12 + 8 = 20 cm, PB = 8 cm, PC = (18 + r) cm, PD = (18 - r) cm

PA × PB = PC × PD

⇒ 8 × 20 = (18 - r) × (18 + r)

⇒ 160 = 324 - r²

⇒ r² = 164

⇒ r = 12.8062 

∴ The radius of the circle is closest to 12.8 cm

Given :

LC = 6, CD = 11, LB = 4 and AB = x 

Formula used:

LC × LD = LB × AL 

Calculations :

According to the question 

LC × LD = LB × AL 

6 × (6 + 11) = 4 × (4 + x) 

⇒ 4 + x = 51/2 

⇒ 4 + x = 25.5 

⇒ x = AB = 21.5 

∴ The length of AB is 21.5 cm.

Given:

AX = 24

XB = k

CX = (k + 2)

XD = 16

Formula Used:

If two chords AB and CD intersecting at point X.

Then, AX × XB = CX × XD

Calculation:

AX × XB = CX × XD

⇒ 24 × k = (k + 2) × 16

⇒ 3k = 2(k + 2)

⇒ 3k - 2k = 4

⇒ k = 4

So, the value of k is 4.

Given:

\(∠ ARS = 125^\circ\)

Concept:

Angle made in a semi-circle is a right angle.

Angles formed in the same segment of a circle will be equal in measure.

Calculation:

BR is formed by joining B and R.

∠ARS + ∠ARP = 180°  [Linear Pair]

⇒ ∠ARP = 180° - 125° = 55° 

∠ARB = 90°    [Angle made in semi-circle] 

⇒ ∠ARP + ∠BRP =  90° 

⇒ ∠BRP = 90° - 55° = 35° 

∠BRP = ∠PAB = 35°  [Angles made in the same segement]

∴ ∠PAB = 35°

Given:

AB = 4 cm and BC = 5 cm

Concept:

Tangent secant segment theorem: If a tangent and secant meet at a common point outside a circle, the segments created have a similar relationship to that of two secant rays.

⇒ AD² = AB (AB + BC)      

Calculation:

Using tangent secant segment theorem, we have,

AD2 = AB (AB + BC)     

⇒ AD2 = 4 (4 + 5)

⇒ AD2 = 36

⇒ AD = 6 cm

Given:

∠BAC = 60°

Concept used:

The angle subtended by an arc of a circle at the center is double the angle subtended by it any point on the remaining part of the circle.

Calculation:

∠BIC = 2 × ∠BAC = 2 × 60° = 120°