NCERT Class 10 Math Chapter 10 Circles Solutions

Here, Class 10 Math Chapter 10 Circles Solutions are provided. Subject matter experts have created these solutions to help students in their board exam. They have created solutions in such a way that students can revise the entire solutions in very little time and can score good marks in the board exam. Every solution is provided in detail and in very simple language.

Class 10 Math Chapter 10 Circles Solutions are very important for board exam. Solving these Circles solutions will boost the confidence and students can score better in exam. Solutions are provided according to latest NCERT syllabus. Experts created these step-by-step solutions, which will definitely be helpful for students.

Circles Solutions

Exercise 10.1

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Exercise 10.2

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Chapter 9: Some Applications of Trigonometry Solution
Chapter 11: Areas Related to Circles Solutions

To understand the chapter thoroughly, it is essential to practice all the questions provided within it, as well as any additional questions. Solving previous years’ question papers also proves to be very helpful in exam preparation.

Important Questions for NCERT Chapter 10: Circles

Question :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) 119 cm.(A)\ 12 \text{ cm} \qquad (B)\ 13 \text{ cm} \qquad (C)\ 8.5 \text{ cm} \qquad (D)\ \sqrt{119} \text{ cm.}

Answer :

(D) 119 cm.(D)\ \sqrt{119} \text{ cm.}

Solution:

In Δ OPQ, angle P is right angle. [Since radius is perpendicular to tangent]\text{In } \Delta \text{ OPQ, angle P is right angle. [Since radius is perpendicular to tangent]}
Using Pythagoras theorem, OQ2=PQ2+OP2⇒122=PQ2+52⇒144=PQ2+25\begin{array}{l} \text{Using Pythagoras theorem, } OQ^2 = PQ^2 + OP^2 \\ \Rightarrow 12^2 = PQ^2 + 5^2 \\ \Rightarrow 144 = PQ^2 + 25 \end{array}
⇒PQ2=144−25=119⇒PQ=119Hence, the option (D) is correct.\begin{array}{l} \Rightarrow PQ^2 = 144 – 25 = 119 \\ \Rightarrow PQ = \sqrt{119} \\ \text{Hence, the option (D) is correct.} \end{array}

Question :

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∘.\textbf{Prove that } \angle AOB = 90^\circ.

Answer :

Let us join point O to C.∠AOB=90∘\begin{array}{l} \text{Let us join point O to C.} \\ \\ \angle AOB = 90^\circ \end{array}

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