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### CBSE Class 10 Maths Chapter 10-Circles Objective Questions

CBSE Class 10 Maths Chapter 10-Circles Objective Questions

The Class 10 Maths Chapter 10 Circles contains details like the properties of circle and the existence of the tangents to a circle. In this chapter, taken from the Unit 4 Geometry of CBSE Syllabus 2019-20, the students are introduced to some complex terms such as tangents, tangents to a circle, number of tangents from a point on the circle and so on. Also, keeping in mind the fact that the number of MCQs are likely to increase for the upcoming board exams, we have compiled here the CBSE Class 10 Maths Chapter 10-Circles Objective Questions.

These CBSE Class 10 Maths objective questions are most likely categorised topic wise.

Find here some of the sub-topics that are discussed via these objective questions we have given below:

10.1 Introduction to Circles (5 MCQs From This Topic)

10.2 Tangent to the Circle (8 MCQs From The Given Topic)

10.3 Theorems (7 MCQs Listed From Topic)

Solution: Let O be the common center of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.

OA 2 =OC 2 +AC 2     [by Pythagoras’ theorem]

Solution: Consider the below figure wherein three circles touch each other externally.

Since the distances between the centres of these circles are 5 cm, 6 cm and 7 cm respectively, we have the following set of equations with respect to the above diagram:

y+z = 6     …… (2) (⇒ y=6-z)…   (2.1)

Adding (1), (2) and (3), we have 2(x+y+z) =5+6+7=18

Using (1) in (4), we have 5+z=9⟹z=4

Therefore, the radii of the circles are 3 cm, 2 cm and 4 cm.

tangent to  a circle is perpendicular to the radius through the point of contact

OA=OC      ——–(radii of the same circle)

→∠OAC=∠OCA=25 ° —– (base angles of an isosceles triangle )

OB=OC      ——–(radii of the same circle)

→∠OBC=∠OCB=35 °  —–(base angles of an isosceles triangle )

∠AOB=2×∠ACB —-(angle at the center is twice the angle at the circumference)

B: Collection of all points equidistant from a fixed point is ______.

Solution: Tangent is a line which touches the circle at only 1 point.

Collection of all points equidistant from a fixed point is called a circle.

Solution: Let O be the centre of the circle and let A be a point outside the circle such that OA = 26 cm.

Let AT be the tangent to the circle.

Since the radius through the point of contact is perpendicular to the tangent, we have ∠OTA = 90°. In right △ OTA, we have

= [(26) 2  – (24) 2 ] = (26 + 24) (26 – 24) = 100.

Hence, the radius of the circle is 10 cm.

ABCD is a cyclic quadrilateral ∴∠A +∠C = 180° and ∠B+ ∠D =  180°

1/2∠A+1/2 ∠C = 90° and 1/2 ∠B+1/2 ∠D =  90°

In △ARB and △CPD, x+y + ∠ARB = 180° and z+w+ ∠CPD =  180°

∠ARB = 180° – (x+y) and ∠CPD = 180° – (z+w)

∠ARB+∠CPD = 360° – (x+y+z+w) = 360° – (90+90)

The sum of a pair of opposite angles of a quadrilateral PQRS is 180∘.

∠OBD = ∠ACD (Angle subtended by the common chord AD)

We know that OB is perpendicular to PQ

(Angle subtended by a chord on the centre is double the angle subtended on the circle)

Consider the triangles OAB and OAC are congruent as

Hence AD = AB/2 = 6/2 = 3 cm as the perpendicular to the chord from the center bisects the chord.

So Area of OAB = 1/2 AB x OD = 1/2 6 x 4 = 12 sq. cm.        ….. (i)

Now AO extended should meet the chord at E and it is middle of the BC as ABC is an isosceles with AB= AC

Triangles AEB and AEC are congruent as

Therefore triangles being congruent, ∠AEB = ∠AEC = 90°

Therefore BE is the altitude of the triangle OAB with AO as base.

Also this implies BE =EC or BC =2BE

= ½×AO×BE = ½ × 5×BE = 12 sq. cm as arrived in eq (i).

Therefore BC = 2BE = 2×4.8 cm = 9.6 cm.

In a circle of radius 17 cm, two parallel chords are drawn on opposite sides of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is

AB = 16 ⇒ AE = BE = 8 cm as perpendicular from centre to the chord bisects the chords

∠OCA = ∠PCB (Vertically opposite angle)

As, tangent to a circle is perpendicular to the line joining the centre of the circle to the tangent at the point of contact to the circle.

Applying Pythagoras theorem to triangle OPQ

∠OAC + ∠OBC + ∠ACB + ∠AOB = 360°​  ….. (sum of angles of a quadrilateral)

90°​​ + 90°​​ + 75°​​ + ∠AOB = 360°

Applying Pythagoras theorem to triangle OPQ

Applying Pythagoras theorem to triangle OAB

Therefore the distance of A from the centre of the circle is 5 cm.

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

These objective questions from the chapter 10 of CBSE Class 10 Textbooks are framed based on the concepts involved in circles. This will help the students to understand the concept better and also for them to score good marks in the examination.