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Circles is one of the shortest chapters in Class 10 maths, and that is the trap. With only two theorems, students treat it as easy, then lose marks on proofs that skip a reason or on a right angle hiding in the figure. This post covers both theorems, how to write their proofs, four worked problems and a practice routine.
What does the Class 10 circles chapter cover?
The CBSE Mathematics curriculum for Class X (2026-27) lists two results under Circles, and asks for both to be proved: the tangent at any point of a circle is perpendicular to the radius through the point of contact, and the lengths of tangents drawn from an external point to a circle are equal. Circles sits in the Geometry unit with triangles, and that unit carries 15 of the 80 theory marks. The NCERT Class 10 chapter on circles, in its 2026-27 reprint, proves both theorems and works through examples. Check the curriculum for your exam year, as CBSE revises it every session.
The chapter starts from three facts. A tangent touches a circle at exactly one point, while a secant cuts it twice. There is no tangent through a point inside the circle, one through a point on it and two from a point outside. And the length of a tangent is the distance from the outside point to the point of contact.
How do you prove the tangent is perpendicular to the radius?
Take a circle with centre O and a tangent XY touching it at P. Pick any other point Q on XY. Q must lie outside the circle, because if it were inside, XY would cut the circle twice and be a secant. So OQ is longer than the radius OP. This is true for every point of XY except P, so OP is the shortest distance from O to the line. The shortest segment from a point to a line is the perpendicular, so OP is perpendicular to XY.
In a board answer, write Given, To prove, Construction and Proof, with a reason beside each step. The reason most often left out is why Q lies outside the circle.
Why are two tangents from an outside point equal?
From an external point P, draw tangents PQ and PR to a circle with centre O. Join OP, OQ and OR. By the first theorem, angles OQP and ORP are both 90°. In right triangles OQP and ORP, OQ = OR (radii of the same circle) and OP is common. So the triangles are congruent by RHS, and PQ = PR by CPCT.
The same congruence shows that OP bisects the angle between the two tangents, which many problems use. NCERT also shows a shorter proof with Pythagoras: PQ² = OP² − OQ² = OP² − OR² = PR².
Four worked examples
Example 1: length of a tangent. A point P is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from P.
Join O to the point of contact T. Angle OTP is 90°, so OP is the hypotenuse. PT² = 13² − 5² = 169 − 25 = 144, which gives PT = 12 cm.
Example 2: angle between tangents. Tangents PA and PB are drawn from P to a circle with centre O, and angle AOB = 110°. Find angle APB.
In quadrilateral OAPB, angles OAP and OBP are 90° each, and the four angles add to 360°. So angle APB = 360° − 90° − 90° − 110° = 70°. In general, the angle between the tangents and the angle at the centre add up to 180°.
Example 3: a quadrilateral around a circle. All four sides of quadrilateral ABCD touch a circle. AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
Let the circle touch AB at P, BC at Q, CD at R and DA at S. Tangents from the same point are equal, so AP = AS, BP = BQ, CR = CQ and DR = DS. Adding, AB + CD = AP + BP + CR + DR = AS + BQ + CQ + DS = AD + BC. So 6 + 4 = AD + 7, and AD = 3 cm.
The same idea proves that a parallelogram drawn around a circle is a rhombus: opposite sides are equal, so AB + CD = AD + BC becomes 2AB = 2BC, and all four sides are equal. That proof appears in the CBSE Mathematics Standard sample paper for 2026-27.
Example 4: perimeter with a third tangent. Tangents PA and PB from P are each 10 cm long. A third tangent touches the circle at C, on the arc between A and B nearer to P, and meets PA at X and PB at Y. Find the perimeter of triangle PXY.
From X, the tangents XA and XC are equal. From Y, YB = YC. So the perimeter is PX + XC + CY + YP = PX + XA + YB + YP = PA + PB = 20 cm. You never need the radius.
What mistakes cost marks in circle questions?
Wrong hypotenuse. The line from the centre to the external point is the hypotenuse, not the tangent.
Right angle at the wrong place. The radius is perpendicular to the tangent only at the point of contact.
Missing right angles in the quadrilateral. In Example 2, forgetting the two 90° angles leaves you stuck.
Assuming what you must prove. In the proof of the second theorem, writing "PQ = PR (tangents from a point)" uses the result itself. You need the congruence.
Unnamed congruence rule. Write RHS, not just "congruent".
If similar triangles appear too, our guide to similarity of triangles in Class 10 covers the criteria. For arcs and sectors, read areas related to circles in Class 10.
A practice routine for this chapter
Day 1: Write both proofs from memory, with Given, To prove, Construction and Proof. Compare with NCERT and underline any step where your reason was missing.
Day 2: Solve the NCERT exercise questions on tangent lengths. Mark every right angle on each figure before you start.
Day 3: Solve three problems with a circle inside a triangle or quadrilateral. List the equal tangent pairs before you add anything.
Day 4: Five mixed questions, including one proof, in 20 minutes. After that, rewrite one proof from memory each week.
If one step still does not make sense, such as why the perpendicular is the shortest distance to a line, the Fermi Concept Library has every concept in mathematics with its own lesson, so you can fix that idea and return to the chapter.
Tonight, draw a circle and an outside point, and prove that the two tangents are equal with every reason written beside its step.
