In RD Sharma Class 10 Chapter 10, students explore important concepts like tangents to a circle, number of tangents, and properties related to tangents from a point. These topics not only strengthen your geometry skills but also help you score well in board exams. That’s why our RD Sharma Class 10 Solutions Chapter 10 Circles are carefully prepared to explain each concept in a clear, step-by-step manner.
These RD Sharma Class 10 Solutions 2025-26 follow the latest CBSE Board pattern and are ideal for exam preparation. Whether you are solving NCERT questions or practicing extra problems, these RD Sharma Circles Class 10 Solutions give you the right guidance. Our NCERT solutions are also helpful for those looking for NCERT Solutions for Class 10 Maths Chapter 10.
Students looking for easy learning can access the RD Sharma Class 10 Circles PDF download, which makes revision faster and learning smoother. All problems from RD Sharma Class 10 Solutions Chapter 10 Circles are solved using simple methods so every student can understand the logic easily.
If you're searching for the best way to understand circles in Class 10, this article is your perfect guide. Keep reading to access the complete RD Sharma Class 10 Circles PDF download.
RD Sharma Class 10 Chapter 10 Circles PDF with Solutions
RD Sharma Class 10 Circles PDF with full solutions can help you a lot. Chapter 10 of RD Sharma is all about tangents, points of contact, and other important topics related to circles. With our RD Sharma Class 10 Solutions 2025-26 PDF, students can understand each concept step-by-step and solve questions easly.
This RD Sharma Class 10 Circles PDF download is made to match the latest CBSE guidelines. It also helps you in exams and homework. The solutions are writen in simple words so every student can learn better and faster. It’s very useful for self-study and doubt clearing.
Download the RD Sharma Class 10 Circles PDF now to revise anytime and anywhere. This guide will make your Maths journey smoother and scoring high in exams more easier.
Important Question of RD Sharma Class 10 Chapter 10: Circles with Solutions
Below are exam-relevant question types from RD Sharma Class 10 Chapter 10 (Circles), covering tangents, tangent lengths, angle properties, cyclic quadrilaterals, and typical constructions, each with a concise solution. Each solution uses standard theorems from the chapter such as “tangent at a point is perpendicular to the radius,” “lengths of tangents from an external point are equal,” “sum of opposite angles in a cyclic quadrilateral is 180°,” and right-triangle Pythagoras, as reflected across authoritative chapter solution references and summaries.
Ques: Fill in the blanks
(i) The common point of a tangent and the circle is called the point of contact.
(ii) A circle may have two parallel tangents.
(iii) A tangent to a circle intersects it in one point.
(iv) A line intersecting a circle in two points is called a secant.
(v) The angle between the tangent at a point on a circle and the radius through the point is 90°.
Solution idea: These are the standard definitions and facts from the chapter introduction.
Ques: How many tangents can a circle have?
Answer: Infinitely many tangents, since there are infinitely many points on the circle.
Ques: Prove: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Proof sketch: Join center O to point of contact P; if a line through P meets the circle at any other point, it is a secant; the limiting position for a secant as the second point approaches P is a tangent, making OP⊥tangent at P by the theorem.
Ques: A circle has center O, radius 8cm. Tangent at A meets a line through O at B with AB=15cm. Find OB.
Solution: OA⊥tangent at A. In right ΔOAB, OB²=OA²+AB²=8²+15²=64+225=289⇒OB=17cm.
Ques: From point P outside a circle, PT is a tangent and P is 13cm from the center O; radius r=5cm. Find PT.
Solution: OP=13, OT=r=5, right ΔOPT gives PT=√(OP²−OT²)=√(169−25)=√144=12cm.
Ques: A point P is 26cm from O; tangent length PT=10cm. Find the radius.
Solution: In right ΔOPT, OT²=OP²−PT²=26²−10²=676−100=576⇒OT=24cm, radius=24cm.
Ques: If PT is tangent at T to a circle with center O, and OP=17cm, OT=8cm, find PT.
Solution: PT=√(OP²−OT²)=√(289−64)=√225=15cm.
Ques: Tangent at P to a circle with center O meets a line through O at Q with PQ=24cm, OQ=25cm. Find the radius.
Solution: OP⊥PQ; OQ²=OP²+PQ² ⇒ 25²=OP²+24² ⇒ OP²=625−576=49 ⇒ OP=7cm.
Ques: From an external point A to a circle with center O, two tangents touch at P and Q. Prove AP=AQ and ∠PAQ=2∠POQ.
Solution: Tangent lengths from an external point are equal, so AP=AQ. Quadrilateral OPAQ has OP=OQ=r and OP⊥AP, OQ⊥AQ; central angle ∠POQ subtends arc PQ, while angle between tangents equals 180°−∠POQ; hence ∠PAQ=180°−∠POQ, and exterior angle relation gives the standard result for angle between tangents vs central angle (or express via angle between tangents equals difference of intercepted arcs); concluding ∠PAQ=180°−∠POQ and also inscribed/central relations yield 2∠POQ+∠PAQ=360°, hence ∠PAQ=2∠POQ (consistent angle chase approach is acceptable).
Ques: In ΔABC, a circle touches sides AB and AC at E and F, and AE is a common external segment. Show that AE is the angle bisector of ∠A if tangents AE from A to the incircle satisfy AE=AF.
Solution: Tangent segments from A to the circle are equal: AE=AF. Similarly, from B and C equal pairs arise. Using equal tangents yields AB−AE=BE and AC−AF=CF with AE=AF, giving BE and CF structured symmetrically; angle bisector theorem follows since the incircle’s touchpoint partitions sides proportionally; or use the property that equal tangents from a vertex imply angle bisected by the internal bisector passing through the incenter.
Ques: In triangle ABC with incircle touching BC at D, AB at E, and AC at F, prove that AE=AF and BE=BD, CF=CD.
Solution: Equal tangents from the same external point to a circle are equal: from A, AE=AF; from B, BE=BD; from C, CF=CD.
Ques: In a circle, chord PQ is parallel to the tangent at R. Prove R bisects arc PRQ.
Solution: Angle between tangent at R and chord RP equals angle in the alternate segment (equal to angle in arc PR), and the parallelism implies symmetry of arcs PR and RQ; hence arcs PR and RQ are equal, so R is the midpoint of arc PRQ.
Ques: Prove: Opposite angles of a cyclic quadrilateral are supplementary.
Solution: Standard theorem: measure of an inscribed angle equals half the measure of its intercepted arc; adding opposite inscribed angles sums to half of 360°, i.e., 180°.
Ques: In the given figure, BCD is tangent to a circle at C and BA is a chord through C. Prove ∠BAC+∠ACD=90°.
Solution: Angle between tangent and chord equals the angle in the alternate segment; thus ∠ACD equals angle in arc AB at point A’s alternate segment, and with right-angle results where radius meets tangent, the sum yields 90° in the stated configuration; formalization uses the tangent-chord theorem plus triangle angle sum.
Ques: Tangents from external points P and Q to two circles intersect at T; common tangents PQ and RS meet at A. Prove PQ=RS (equal lengths of common tangents).
Solution: For two circles with a pair of direct common tangents, homothety about the external center maps one circle to the other and preserves tangent length segments between touchpoints, giving equality; or use similar triangles formed by radii to touchpoints and the intersection of tangents.
Ques: If a quadrilateral ABCD circumscribes a circle (touches the circle on all four sides), prove AB+CD=BC+AD.
Solution: Use equal tangent segments from each vertex to the points of contact: From A, AE=AF; from B, BE=BG; from C, CF=CH; from D, DG=DH; summing suitably yields opposite side sums equal.
Ques: Two concentric circles with outer radius 5cm have a chord of the outer circle AC=8cm that is tangent to the inner circle. Find the radius of the inner circle.
Solution: Let O be center; distance from O to chord AC is d=√(R²−(AC/2)²)=√(25−16)=3cm. Since inner circle is tangent to AC, its radius equals this perpendicular distance, so r=3cm.
Ques: If from any point on the common chord of two intersecting circles tangents are drawn to both circles, prove those tangents have equal lengths.
Solution: A point on the common chord has equal power with respect to both circles; length of tangent squared equals power of the point, hence equal for both circles; thus tangents are equal.
Ques: A chord and a tangent form an angle θ at the point of contact. Show that θ equals the angle in the alternate segment.
Solution: This is the tangent-chord theorem: angle between a tangent and chord through the point of contact equals the angle in the alternate segment of the circle subtended by the chord.
Ques: In ΔABC right-angled at B with AB=8cm, BC=6cm, find the inradius.
Solution: For right triangle, inradius r=(a+b−c)/2 where c=AC=√(8²+6²)=10. Then r=(8+6−10)/2=4/2=2cm.
RD Sharma Class 10 Solutions Chapter 10: Circles Overview
RD Sharma Class 10 Chapter 10 – Circles helps students learn important circle ideas in a simple way, matching the latest CBSE Class 10 syllabus and exam style. This chapter mainly focuses on tangents, secants, chords, and key theorems used to solve questions fast and correctly.
What this chapter teaches
- Understanding circle terms: centre, radius, diameter, chord, secant, tangent, point of contact.
- Tangent basics: a tangent touches the circle at exactly one point; tangent is perpendicular to the radius at the point of contact.
- Lengths of tangents from an external point are equal (PQ = PR).
- Number of tangents from a point and parallel tangents concepts.
- Angle facts and properties used in problem-solving and proofs (e.g., cyclic ideas appear in exercises).
- Exercise-wise practice with step-by-step solved examples to build accuracy and speed.
Why use RD Sharma Solutions for Chapter 10 – Circles
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Clear, step-by-step solutions in simple language following CBSE patterns.
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Helpful for homework, tests, and board exam preparation with lots of solved examples and practice.
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Important theorems highlighted:
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The tangent at any point of a circle is perpendicular to the radius.
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Tangents from an external point to a circle are equal in length.
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Exercise coverage: two main exercises focusing on tangent properties, number/length of tangents, and related theorems.
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Available as PDF to study online or offline; good for revision and last-minute prep.
Advantages of solving Chapter 10 (Circles)
- Builds strong concepts: Understand secant vs tangent, chord vs diameter, and how perpendicular radius helps in questions.
- Improves problem-solving: Regular practice boosts geometry reasoning and RHS congruence applications in tangent problems.
- Board-exam ready: Questions match CBSE Class 10 style; many direct theorem-based items appear in exams.
- Increases speed and accuracy: Repeated Pythagoras/tangent-length problems make calculations faster.
- Full solutions: Each step explained to avoid getting stuck and to learn the method, not just the answer.
- Helpful for higher studies and competitive basics: Circle theorems are used later in geometry and aptitude tests.