NCERT Solutions for Class 10 Maths Unit 10

Circles Class 10

Unit 10 Circles Exercise 10.1, 10.2 Solutions

A Circle is a simple shape of Euclidean geometry that is the set of all points in a plane that are at a given distance from a given point, the centre. The distance between any of the points and the centre is called the radius. It can also be defined as the locus of a point equidistant from a fixed point.

A circle is a simple closed curve which divides the plane into two regions: an interior and an exterior. In everyday use, the term "circle" may be used interchangeably to refer to either the boundary of the figure, or to the whole figure including its interior; in strict technical usage, the circle is the former and the latter is called a disk.
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Exercise 10.1 : Solutions of Questions on Page Number : 204

Q1 :  

How many tangents can a circle have?


Answer :

A circle can have infinite tangents.

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Q2 :  

Fill in the blanks:

(i) A tangent to a circle intersects it in _______ point (s).

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

(iii) A circle can have __________ parallel tangents at the most.

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


Answer :

(i) One

(ii) Secant

(iii) Two

(iv) Point of contact

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Q3 :  

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 so that OQ = 12 cm. Length PQ is :

(A) 12 cm. (B) 13 cm (C) 8.5 cm (D) cm


Answer :

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Q4 :  

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.


Answer :

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Exercise 10.2 : Solutions of Questions on Page Number : 213

Q1 :  

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(A) 7 cm (B) 12 cm

(C) 15 cm (D) 24.5 cm


Answer :

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Q2 :  

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110, then ∠PTQ is equal to

(A) 60 (B) 70

(C) 80 (D) 90


Answer :

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Q3 :  

If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80, then ∠POA is equal to

(A) 50 (B) 60

(C) 70 (D) 80


Answer :

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Q4 :  

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


Answer :

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Q5 :  

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


Answer :

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Q6 :  

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.


Answer :

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Q7 :  

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.


Answer :

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Q8 :  

A quadrilateral ABCD is drawn to circumscribe a circle (see given figure) Prove that AB + CD = AD + BC


Answer :

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Q9 :  

In the given figure, 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.


Answer :

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Q10 :  

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.



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Q11 :  

Prove that the parallelogram circumscribing a circle is a rhombus.


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Q12 :  

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.


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Q13 :  

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.


Answer :

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<< Previous Chapter 9 : Some Applications of Trigonometry Next Chapter 11 : Constructions >>

Maths : CBSE NCERT Exercise Solutions for Class 10th for Circles ( Exercise 10.1, 10.2 ) will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

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