Circles Class 10 Maths Notes

Circle geometry used to be the longest chapter in Class 9 and the start of Class 10. The rationalised 2024-25 syllabus keeps only the tangent results, so this is now a short chapter with very high yield. Two theorems, one numerical pattern, and a definition, and that is the whole examinable content.Read the chapter as two halves. The first half builds the one fact about a single tangent — it stands at right angles to the radius. The second half uses that fact twice to show that two tangents from the same external point have equal lengths.

• Retained and assessed — the tangent to a circle at the point of contact is perpendicular to the radius through the point of contact, to be proved.
• Retained and assessed — the lengths of the tangents drawn from an external point to a circle are equal, to be proved.
• Everything else from the old chapter — equal chords subtend equal angles at the centre, the perpendicular from the centre bisects a chord, angles in the same segment, the angle at the centre is twice the angle at the circumference, and cyclic quadrilaterals — is out of scope this year.
• The entire Constructions chapter is deleted, so there is no construction question involving tangents anywhere in the paper.

In every diagram in this chapter you will see three named points. O is the centre of the circle. A is the point of contact, where the tangent touches. P is the external point from which the tangent is drawn. Then OA is a radius and PA is a tangent segment, and the angle OAP between them is the angle that the two theorems are about.

• O — the centre of the circle. Every segment from O to the circle is a radius and all of them are equal.
• A — the point of contact, the single point where the tangent touches the circle.
• P — a point outside the circle, from which one or more tangents are drawn.
• PA and PB — the two tangent segments from P, touching the circle at A and at B.
• Line OP — joins the external point to the centre, and it is the hypotenuse of every right triangle in this chapter.

Statement: the tangent to a circle at a point is perpendicular to the radius through the point of contact. This is the first theorem and the second is built on it, so write it out in full with the reason at each step. A five-mark question on this theorem is won or lost on the reasons.

• Given — a circle with centre O, a point A on the circle, and PA a tangent to the circle at A.
• To prove — PA ⟂ OA.
• Proof step 1 — OP and OA are radii of the same circle, so OP = OA. Reason: radii of a circle are equal.
• Proof step 2 — in triangle OAP, ∠OPA = ∠OAP. Reason: angles in an isosceles triangle standing on equal sides OP and OA are equal.
• Proof step 3 — ∠OPA = ∠OAP = ∠AOP. Reason: angles in a triangle add to 180°, so the two equal angles together with the third fill 180°.
• Proof step 4 — ∠AOP = 90°. Reason: three times ∠AOP = 180°.
• Conclusion — therefore PA ⟂ OA, since angles standing on the same line PA are equal to one another.

Statement: the lengths of the tangents drawn from an external point to a circle are equal. So if PA and PB are the two tangents from P, touching the circle at A and at B, then PA = PB. The proof is a three-step congruence, and Theorem 1 is used twice to get the right angles.

• Given — a circle with centre O, PA and PB the tangents from an external point P, touching at A and B.
• To prove — PA = PB.
• Step 1 — OA = OB, since both are radii of the same circle.
• Step 2 — ∠OAP = ∠OBP = 90°, by Theorem 1, since each tangent stands at right angles to the radius at the point of contact.
• Step 3 — OP = OP, the side common to both triangles.
• Conclusion — therefore ΔOAP ≅ ΔOBP by RHS congruence, right angle, hypotenuse and side. Hence by CPCT, the corresponding sides are equal, so PA = PB.

Almost every numerical in this chapter is one Pythagoras step once Theorem 1 has been used. The right angle always sits at the point of contact, and the hypotenuse is always the line from the external point to the centre. Note what that step is and is not: you are not recalling a Class 10 theorem, because both Pythagoras results are deleted. Theorem 1 gives you the right angle, and then the Class 9 relation you already know finishes the arithmetic.Example — a circle has centre O and radius 7 cm. From an external point P, PA is drawn as a tangent to the circle at A. If OP = 25 cm, find the length of PA.

• OA = 7 cm, since OA is a radius of the circle.
• ∠OAP = 90°, by Theorem 1, since PA is a tangent at the point of contact A.
• So OP is the hypotenuse of the right triangle OAP, and OP² = OA² + PA².
• Putting in the values, 25² = 7² + PA², so 625 = 49 + PA².
• Therefore PA² = 625 − 49 = 576, and PA = √576 = 24 cm.
• Answer: the length of the tangent segment PA is 24 cm.

A very common mistake is to file circle geometry under Mensuration because both chapters talk about circles. They are not in the same unit. Circles is the opening topic of Unit IV Geometry, which carries 15 of the 80 theory marks. The area of a circle, the area of a sector and the area of a segment are all in Unit VI Mensuration and carry 10 marks between them.

• Unit I — Number Systems, 6 marks.
• Unit II — Algebra, 20 marks.
• Unit III — Coordinate Geometry, 6 marks.
• Unit IV — Geometry, 15 marks, and circle geometry sits here.
• Unit V — Trigonometry, 12 marks.
• Unit VI — Mensuration, 10 marks, which holds the areas related to circles.
• Unit VII — Statistics and Probability, 11 marks.
• Internal assessment is 20 marks, split 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical.

The bridge to similarity is one line. Because PA = PB, the two triangles you get from a tangent and a secant through P — ΔPQA and ΔPAR — are similar by SAS, and from that similarity the tangent relation PA² = PQ × PR follows. The tangent relation itself is not asked this year, but seeing it once explains why tangents turn up in later work on similarity of triangles.Because the chapter is short, the paper leans on MCQs and assertion-reason, where each theorem can be tested in a single line.

• State the theorem: the tangent at a point of contact is perpendicular to the radius through it.
• Give a reason in each step of the proof of ∠OAP = 90°.
• In ΔOAP, OA = 5 cm and OP = 13 cm. Find PA. Answer 12 cm.
• State and prove that the tangents drawn from an external point to a circle are equal.
• Assertion-reason: a tangent at the point of contact bisects nothing, but it is perpendicular to the radius — assert and give the reason.
• MCQ trap: a tangent to a circle at a point cuts the circle at that one point only, so it has exactly one point in common with the circle, not two and not none.