Circles
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
Class 10 Circles revision notes cover tangents to a circle: the tangent–radius perpendicular theorem, equal tangent lengths from an external point, and their use with triangles and quadrilaterals.
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This chapter is about lines that touch a circle at one point. These lines are called tangents. The central fact is that the radius drawn to the point of contact is perpendicular to the tangent. Once this fact is accepted, every tangent problem contains a right triangle, so distances are usually found with Pythagoras' theorem. The number of tangents that can be drawn from a point also matters: no tangent from a point inside the circle, exactly one tangent at a point on the circle, and exactly two tangents from an external point.
The other big theorem says that the two tangent segments from an external point to the points of contact are equal. It is proved by comparing the two right triangles formed with the common line from the centre to the external point. The same comparison also shows that this line bisects the angle between the tangents. Equality of tangent segments is the tool that joins tangents with triangles and quadrilaterals, because each vertex of a circumscribed figure can be marked with equal tangent lengths. The sums of opposite sides of a quadrilateral circumscribing a circle then come out equal.
Numerical problems usually supply a radius and a distance from the centre, or give one angle and ask for another. A separate family of questions involves circles touching the sides of a triangle or quadrilateral. The safest habit is to draw the radii to the points of contact first and write 90° at each contact point. Mistakes occur when students treat any line meeting the circle twice as a tangent, or forget that a point inside a circle has no tangent. Focus on the point of contact, the right angle there, and equal tangent lengths.
Key terms
- Tangent to a circle
- A tangent is a line that touches the circle at exactly one point. If a line cuts the circle at two points, it is not a tangent.
- Point of contact
- The single point at which a tangent meets the circle. The radius drawn to this point is perpendicular to the tangent.
- Number of tangents from a point
- From an interior point, no tangent can be drawn. From a point on the circle, exactly one tangent exists. From an external point, exactly two tangents exist.
- Tangent–radius perpendicular theorem
- At the point of contact, the tangent is perpendicular to the radius through that point. Conversely, a line through a point of the circle perpendicular to the radius at that point is a tangent.
- Equal tangent lengths
- The two tangent segments drawn from the same external point to the circle are equal in length.
- Right triangle formed by a tangent
- For an external point P, contact point A and centre O, the angle at A is 90°, so triangle OAP is right-angled at A. This allows lengths to be found by Pythagoras' theorem.
- Tangential triangle or quadrilateral
- A triangle or quadrilateral is called circumscribed when all its sides touch a circle. Equal tangent lengths from each vertex give side relationships such as AB + CD = BC + AD for a circumscribed quadrilateral.
- Line from external point to centre
- The line joining an external point P to the centre O bisects the angle between the two tangents drawn from P. It follows from the congruence of the two right triangles formed.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Length of a tangent from an external point | t = √(OP² − r²) | t is the length of the tangent segment from external point P to the point of contact, OP is the distance from P to the centre O, and r is the radius of the circle. It is valid only when P is outside the circle, so OP > r. |
| Angle between radii and angle between tangents | ∠AOB + ∠APB = 180° | A and B are the contact points of the two tangents drawn from external point P, and O is the centre. It holds because ∠OAP = ∠OBP = 90° and the angles of a quadrilateral sum to 360°. |
| Opposite sides of a circumscribed quadrilateral | AB + CD = BC + AD | ABCD is a quadrilateral whose four sides all touch a circle. Each letter pair stands for the length of that side. The relation follows from equality of tangent segments from each vertex. |
Practice questions with answers
1. State how many tangents to a circle can be drawn from a point inside it, from a point on it, and from a point outside it.
No tangent can be drawn from a point inside the circle, because every line through such a point cuts the circle at two points. From a point on the circle exactly one tangent exists. From an external point exactly two tangents exist.
2. The tangent at point A of a circle with centre O meets a line through O at B. If OA = 8 cm and AB = 15 cm, find OB.
Since AB is the tangent at A, OA is perpendicular to AB. Therefore triangle OAB is right-angled at A. Using Pythagoras' theorem, OB = √(8² + 15²) = √(64 + 225) = √289 = 17 cm.
3. Prove that the tangent segments drawn from an external point to a circle are equal.
Let P be the external point and PA, PB be tangents touching the circle at A and B, with centre O. Then OA = OB, OP is common, and ∠OAP = ∠OBP = 90°. Triangles OAP and OBP are congruent by RHS congruence, so PA = PB.
4. From an external point P, two tangents PA and PB touch a circle with centre O at A and B. If ∠APB = 60°, find ∠AOB.
Since PA⊥OA and PB⊥OB, ∠OAP = 90° and ∠OBP = 90°. In quadrilateral OAPB, the angle sum is 360°, so ∠AOB = 360° − (90° + 90° + 60°) = 120°.
5. A circle touches the sides AB, BC and CA of triangle ABC at D, E and F respectively. If AF = 6 cm, BD = 4 cm and CE = 5 cm, find the three sides of the triangle.
Tangent segments from the same vertex are equal, so AD = AF = 6 cm, BE = BD = 4 cm and CF = CE = 5 cm. Therefore AB = 6 + 4 = 10 cm, BC = 4 + 5 = 9 cm, and CA = 5 + 6 = 11 cm.
6. A quadrilateral ABCD circumscribes a circle. If AB = 4 cm, BC = 6 cm and CD = 7 cm, find AD.
For a circumscribed quadrilateral, AB + CD = BC + AD. Substituting the given lengths gives 4 + 7 = 6 + AD, so 11 = 6 + AD and AD = 5 cm.
7. Tangents TP and TQ are drawn from point T to a circle with centre O, touching at P and Q. If ∠PTQ = 70°, find ∠OPQ.
OP⊥TP and OQ⊥TQ, so ∠OPT = 90° and ∠OQT = 90°. In quadrilateral O-P-T-Q, ∠POQ + 90° + 70° + 90° = 360°, giving ∠POQ = 110°. Since OP = OQ, triangle OPQ is isosceles, so ∠OPQ = (180° − 110°)/2 = 35°.
8. Find the length of a tangent drawn from a point 26 cm away from the centre of a circle of radius 10 cm.
Let P be the external point and T the point of contact. Since the radius is perpendicular to the tangent, triangle OPT is right-angled at T. Therefore PT = √(OP² − OT²) = √(26² − 10²) = √(676 − 100) = √576 = 24 cm.
9. Can a tangent be drawn from a point that is 4 cm from the centre of a circle of radius 5 cm? Give a reason.
No. The point lies inside the circle because its distance from the centre, 4 cm, is less than the radius, 5 cm. Any line drawn through an interior point cuts the circle at two points, so it cannot be a tangent.
10. Prove that the tangents drawn at the endpoints of a diameter of a circle are parallel.
Let AB be a diameter of the circle, and let tangents be drawn at A and B. Each tangent is perpendicular to the radius at its point of contact, so both tangents are perpendicular to the same line AB. Two lines perpendicular to the same line are parallel, hence the tangents are parallel.
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