Is it correct to say: Using the wrong hypotenuse, e.g. writing PA² = OP² + r².?

The right angle is at the point of contact A, so OP is the hypotenuse: PA² = OP² − OA².

Is it correct to say: Taking the angle between the tangents to be equal to the angle at the centre.?

They are supplementary: ∠APB + ∠AOB = 180°. If ∠APB = 70°, then ∠AOB = 110°.

Is it correct to say: Writing AB + BC = CD + DA for a quadrilateral circumscribing a circle.?

It is the OPPOSITE sides whose sums are equal: AB + CD = AD + BC.

Is it correct to say: Proving the equal-tangent theorem with SAS, or without giving the right angle a reason.?

Triangles OAP and OBP are right-angled at A and B (tangent ⟂ radius), with OA = OB and OP common. The rule is RHS.

Is it correct to say: Using the full angle between the tangents in a trigonometry step.?

OP bisects ∠APB. For tangents inclined at 60°, ∠APO = 30°, so sin 30° = OA/OP.

Is it correct to say: Assuming a tangent can be drawn from a point inside the circle.?

From an interior point there is no tangent; from a point on the circle there is one; from an exterior point there are exactly two.

📄 Formula sheet (PDF)

Circles

Tangent to a circle, tangent perpendicular to the radius at the point of contact, equal tangents from an external point and their applications — NCERT Class 10 Maths Ch 10

📐 14 formulas✏️ 4 examples🎯 6 practice⚖️ Unit IV (Geometry): 15 marks (with Ch 6)🏫 CBSE📚 Class 10✓ 2025–26 syllabus

💡

Board Exam Tips

→Both theorems (tangent ⟂ radius, and equal tangents from an external point) are proofs in the CBSE syllabus. Learn each one with its labelled figure.
→In any tangent problem, first join the centre to the point of contact. You get a right angle, and usually a right triangle for Pythagoras.
→From an external point P, the two tangents are equal and OP bisects both the angle between the tangents and the angle between the radii.
→For a quadrilateral or triangle drawn around a circle, mark equal tangent lengths from each vertex with the same letter. The algebra then becomes short.
→The angle between the two tangents and the angle between the radii at the centre always add up to 180°.
→Write the reason next to every step of a proof: 'tangent ⟂ radius', 'tangents from an external point are equal', 'RHS congruence'.

📐 Formulas(14)

Line and Circle: Three Cases

common points = 0 (non-intersecting), 1 (tangent), 2 (secant)

Number of Tangents from a Point

P inside: 0, P on the circle: 1, P outside: 2

Tangent Perpendicular to Radius (Theorem 10.1)★ Board fav

O A ⊥ X Y, ∠ O A X = ∠ O A Y = 9 0^{\circ}

Symbol	Meaning
$O$	Centre of the circle
$A$	Point of contact
$X Y$	Tangent at A

Length of a Tangent★ Board fav

P A = O P^{2} - r^{2}

Symbol	Meaning
$P A$	Length of the tangent from P to the point of contact A (cm)
$O P$	Distance of the external point P from the centre O (cm)
$r$	Radius, r = OA (cm)

Equal Tangents from an External Point (Theorem 10.2)★ Board fav

P A = P B

Symbol	Meaning
$P$	External point
$A, B$	Points of contact of the two tangents

Centre Lies on the Angle Bisector

∠ A P O = ∠ B P O, ∠ A O P = ∠ B O P

Angle Between Tangents and Angle at Centre★ Board fav

∠ A P B + ∠ A O B = 18 0^{\circ}

Angle Between Tangents and the Chord of Contact

∠ A P B = 2 ∠ O A B

OP Is the Perpendicular Bisector of the Chord of Contact

O P ⊥ A B, A M = M B

Tangents at the Ends of a Diameter

tangent at A ∥ tangent at B (A B a diameter)

Chord of the Larger of Two Concentric Circles

A B = 2 R^{2} - r^{2}

Symbol	Meaning
$R$	Radius of the larger circle (cm)
$r$	Radius of the smaller circle (cm)

Quadrilateral Circumscribing a Circle

A B + C D = A D + B C

Angles at the Centre for a Circumscribed Quadrilateral

∠ A O B + ∠ C O D = 18 0^{\circ}, ∠ B O C + ∠ A O D = 18 0^{\circ}

Triangle Circumscribing a Circle

A B = x + y, B C = y + z, C A = z + x, ar (△ A B C) = r \cdot s

Symbol	Meaning
$x, y, z$	Tangent lengths from vertices A, B, C (cm)
$r$	Radius of the inscribed circle (cm)
$s$	Semi-perimeter of the triangle (cm)

✏️ Solved Examples

1Solved Exampleeasy3 steps

A point P is 17 cm from the centre O of a circle of radius 8 cm. Find the length of the tangent drawn from P to the circle.

Let the tangent touch the circle at A. OA ⟂ PA (tangent ⟂ radius), so triangle OAP is right-angled at A with hypotenuse OP.

P A^{2} = O P^{2} - O A^{2}

2Solved Exampleboard4 steps

PA and PB are tangents from an external point P to a circle with centre O. If ∠APB = 50°, find ∠AOB, ∠OAB and ∠OPA.

∠OAP = ∠OBP = 90°, so in quadrilateral OAPB the angles at P and O are supplementary.

∠ A O B = 18 0^{\circ} - 5 0^{\circ} = 13 0^{\circ}

3Solved Exampleboard3 steps

A quadrilateral ABCD is drawn to circumscribe a circle. If AB = 8 cm, BC = 9 cm and CD = 6 cm, find AD.

Tangents from each vertex are equal, which gives the result that opposite sides have equal sums.

A B + C D = A D + B C

4Solved ExampleHOTS5 steps

AB is a chord of length 24 cm of a circle with centre O and radius 15 cm. The tangents at A and B meet at P. Find the length PA.

PA = PB and OA = OB, so OP is the perpendicular bisector of AB. Let OP meet AB at M; then AM = 12 cm and ∠OMA = 90°.

O M = O A^{2} - A M^{2} = 1 5^{2} - 1 2^{2} = 81 = 9 cm

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using the wrong hypotenuse, e.g. writing PA² = OP² + r².
✓The right angle is at the point of contact A, so OP is the hypotenuse: PA² = OP² − OA².
2
Taking the angle between the tangents to be equal to the angle at the centre.
✓They are supplementary: ∠APB + ∠AOB = 180°. If ∠APB = 70°, then ∠AOB = 110°.
3
Writing AB + BC = CD + DA for a quadrilateral circumscribing a circle.
✓It is the OPPOSITE sides whose sums are equal: AB + CD = AD + BC.
4
Proving the equal-tangent theorem with SAS, or without giving the right angle a reason.
✓Triangles OAP and OBP are right-angled at A and B (tangent ⟂ radius), with OA = OB and OP common. The rule is RHS.
5
Using the full angle between the tangents in a trigonometry step.
✓OP bisects ∠APB. For tangents inclined at 60°, ∠APO = 30°, so sin 30° = OA/OP.
6
Assuming a tangent can be drawn from a point inside the circle.
✓From an interior point there is no tangent; from a point on the circle there is one; from an exterior point there are exactly two.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
The length of the tangent from a point Q to a circle is 24 cm, and Q is 26 cm from the centre. Find the radius of the circle.
Q2
PA and PB are tangents from P to a circle with centre O. If ∠AOB = 115°, find ∠APB.
Q3
Two concentric circles have radii 13 cm and 5 cm. Find the length of the chord of the larger circle that touches the smaller circle.
Q4
A circle touches all four sides of a quadrilateral PQRS. If PQ = 7 cm, QR = 10 cm and RS = 9 cm, find SP.
Q5
The incircle of ΔABC touches BC, CA and AB at D, E and F. If AB = 10 cm, BC = 12 cm and CA = 8 cm, find AF, BD and CE.
Q6
Two tangents PA and PB from P to a circle of radius 6 cm are inclined to each other at 60°. Find OP and PA.

📝 Notes

Circles

This chapter studies one line: the tangent, which touches a circle at exactly one point. Two theorems, both proofs in the syllabus, drive every question.

The two theorems

Theorem 10.1: the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Theorem 10.2: the lengths of the two tangents drawn from an external point to a circle are equal.

Theorem 10.2 is proved using Theorem 10.1 (right angles at the points of contact) and RHS congruence, so learn them in that order.

A routine for tangent questions

Join the centre O to every point of contact. Mark each angle there as 90°.
Join O to the external point P. Triangle OAP is right-angled at A, so $PA^2 = OP^2 - r^2$ .
Use PA = PB. Triangle PAB is isosceles, and OP bisects ∠APB and ∠AOB.
Use angle sums: in quadrilateral OAPB, ∠APB + ∠AOB = 180°.

Most numerical tangent questions use only these four steps, with Pythagoras or a trigonometric ratio at the end.

Polygons drawn around a circle

When every side of a triangle or quadrilateral touches the circle, split each side at its point of contact. Tangent lengths from the same vertex are equal, so give them one letter. For a quadrilateral this gives AB + CD = AD + BC at once. For a triangle with tangent lengths x, y, z from A, B, C, the sides are x + y, y + z and z + x, and the semi-perimeter is x + y + z.

Checks before you finish

A tangent length must be shorter than OP, the distance from the centre to the external point.
The angle between the tangents is always less than 180°, and the angle at the centre makes it up to 180°.