Theorems on Tangents to a Circle

A tangent is a straight line that touches a circle at exactly one point. The tangent, radius, centre and point of contact satisfy some very important geometric relationships. These relationships are used to construct tangents, find unknown angles, prove equal lengths and study two circles that touch each other.

This page explains the three main tangent theorems and their important applications step by step.

📚 Tangent and point of contact

Suppose a line AB touches a circle with centre O only at the point P.

👉 AB is a tangent to the circle.

👉 P is the point of contact.

👉 OP is the radius through the point of contact.

👉 A point lying outside the circle is called an external point.

⭐ Theorem 40: The tangent at any point of a circle is perpendicular to the radius through the point of contact

Given:	AB is a tangent at P to a circle with centre O, and OP is the radius through the point of contact.
To prove:	OP ⟂ AB.
Construction:	Take any point Q on AB other than P and join OQ. Let OQ meet the circle at R.
Proof:	Since Q is a point on the tangent other than the point of contact P, Q lies outside the circle. Therefore R lies between O and Q.
∴	OR < OQ. [R lies between O and Q]
Also,	OR = OP. [Radii of the same circle]
∴	OP < OQ.
Thus,	Among all line segments drawn from O to points of the line AB, OP is the shortest.
∴	The shortest distance from a point to a line is the perpendicular distance. Hence OP ⟂ AB. [Proved]

📚 Direct meaning of the theorem

The angle between a tangent and the radius through the point of contact is always a right angle. Therefore,

∠OPA = ∠OPB = 90°.

This result is the basic idea behind most proofs involving tangents.

⭐ Converse: A line drawn perpendicular to a radius at its endpoint on the circle is a tangent to the circle

Given:	OP is a radius of a circle with centre O, and a line AB is drawn through P such that OP ⟂ AB.
To prove:	AB is a tangent to the circle at P.
Construction:	Imagine or draw the tangent CD to the circle at P.
Proof:	Since CD is the tangent at P and OP is the radius through the point of contact, OP ⟂ CD.
∴	∠OPD = 90°.
Also,	By the given condition, OP ⟂ AB, so ∠OPB = 90°.
∴	∠OPD = ∠OPB.
Therefore,	CD and AB are the same straight line because only one perpendicular can be drawn to a given line at a given point.
∴	AB is the tangent to the circle at P. [Proved]

📚 Application 1: Only one tangent can be drawn at a fixed point on a circle

Let P be a fixed point on a circle and let OP be the radius through that point.

By the tangent theorem, the tangent at P must be perpendicular to OP. But through P, only one perpendicular line can be drawn to OP.

Therefore, exactly one tangent can be drawn at a fixed point on a circle.

📚 Application 2: The perpendicular to a tangent at the point of contact passes through the centre

Let AB be a tangent to a circle at P. Draw PC perpendicular to AB at P.

By the tangent theorem, the radius OP is also perpendicular to AB. Thus,

OP ⟂ AB and PC ⟂ AB.

Only one perpendicular to AB can pass through P. Hence PC and PO are the same straight line.

Therefore, the perpendicular drawn to a tangent at the point of contact passes through the centre of the circle.

📚 Application 3: Finding angles using a tangent and a line through the centre

Let AT be a tangent at A to a circle with centre O. Let B, O, T be collinear and let OA and OB be radii. Suppose

∠ATO = x° and ∠OAB = ∠OBA = y°.

We will prove that x = 90 - 2y.

Since,	AT is tangent at A and OA is the radius through the point of contact,
∴	∠OAT = 90°.
Also,	∠ATO = x°.
∴	In △AOT, ∠AOT = 90° - x°. ...(i)
Again,	OA = OB. [Radii of the same circle]
∴	∠OAB = ∠OBA = y°.
Now,	The exterior angle of △AOB gives ∠AOT = ∠OAB + ∠OBA = 2y°. ...(ii)
∴	From (i) and (ii), 2y° = 90° - x°.
Therefore,	x = 90 - 2y.

💡 Example

If y = 25°, then

x = 90° - 2 × 25° = 40°.

📚 Application 4: Two tangents can be drawn from an external point to a circle

Let T be a point outside a circle with centre O. We show that two tangents can be drawn from T to the circle.

Construction:	Join T to O. Draw a circle with TO as diameter. Let it meet the given circle at A and B. Join TA, TB, OA and OB.
Proof:	In the circle with diameter TO, both ∠OAT and ∠OBT are angles in a semicircle.
∴	∠OAT = ∠OBT = 90°.
That is,	OA ⟂ AT and OB ⟂ BT.
Also,	OA and OB are radii through A and B.
∴	By the converse of the tangent theorem, TA and TB are tangents to the given circle at A and B respectively.
Hence,	two tangents can be drawn from the external point T. [Proved]

⭐ Theorem 41: Tangents drawn from the same external point to a circle are equal in length

The line joining the centre to the external point also forms equal angles with the two radii drawn to the points of contact.

Given:	From an external point P, two tangents PA and PB are drawn to a circle with centre O. Their points of contact are A and B. Join OA, OB and OP.
To prove:	(i) PA = PB and (ii) ∠POA = ∠POB.
Proof:	PA and PB are tangents, while OA and OB are radii through their points of contact.
∴	OA ⟂ PA and OB ⟂ PB.
∴	∠OAP = ∠OBP = 90°.
Now,	In right triangles △PAO and △PBO, OP is the common hypotenuse and OA = OB. [Radii of the same circle]
∴	△PAO ≅ △PBO. [RHS congruence]
∴	PA = PB. [Corresponding sides of congruent triangles] [(i) Proved]
And,	∠POA = ∠POB. [Corresponding angles of congruent triangles] [(ii) Proved]

📚 Another important result

From the same congruence, we also get

∠APO = ∠OPB.

Therefore, OP bisects both the central angle ∠AOB and the angle ∠APB between the two tangents.

💡 Simple example

From an external point P, tangents PA and PB are drawn to a circle. If PA = 7 cm, then

PB = 7 cm.

If ∠POA = 35°, then

∠POB = 35°.

⭐ Theorem 42: If two circles touch each other, their point of contact lies on the line joining their centres

This theorem is true for both external contact and internal contact.

Given:	Two circles with centres A and B touch each other at P.
To prove:	A, P and B are collinear.
Construction:	Join AP and BP. Draw the common tangent ST to the two circles at P.
Proof:	For the circle with centre A, ST is tangent at P and AP is the radius through the point of contact.
∴	AP ⟂ ST.
Similarly,	For the circle with centre B, ST is tangent at P and BP is the radius through the point of contact.
∴	BP ⟂ ST.
Therefore,	AP and BP are both perpendicular to ST at the same point P. Only one perpendicular can be drawn to a line at a given point.
∴	AP and BP lie on the same straight line. Hence A, P and B are collinear. [Proved]

📚 Distance between centres in external and internal contact

Let the radii of the two circles be r₁ and r₂.

👉 External contact: The point of contact P lies between the two centres. Therefore,

AB = AP + PB = r₁ + r₂.

👉 Internal contact: The two centres lie on the same side of the point of contact. If the larger radius is R and the smaller radius is r, then

AB = R - r.

💡 More solved examples

Example 1: Length of a tangent

From an external point P, two tangents PA and PB are drawn to a circle. If PA = 12 cm, find PB.

Since tangents from the same external point are equal,

PB = PA = 12 cm.

Example 2: Tangent length using Pythagoras theorem

A circle has centre O and radius OA = 5 cm. From an external point P, PA is tangent and OP = 13 cm. Find PA.

Since OA ⟂ PA, △OAP is right-angled at A.

OP² = OA² + PA² 13² = 5² + PA² PA² = 169 - 25 = 144

PA = 12 cm.

Example 3: Distance between centres of touching circles

Two circles with radii 6 cm and 4 cm touch externally. The distance between their centres is

6 + 4 = 10 cm.

If the same circles touch internally, the distance between their centres is

6 - 4 = 2 cm.

✍️ Practice problems

A tangent PT touches a circle with centre O at T. Find ∠OTP.
From an external point P, tangents PA and PB are drawn to a circle. If PA = 9.5 cm, find PB.
A circle has radius 7 cm. A tangent of length 24 cm is drawn from an external point P. Find the distance from P to the centre.
Two circles of radii 8 cm and 5 cm touch externally. Find the distance between their centres.
Two circles of radii 9 cm and 4 cm touch internally. Find the distance between their centres.
From a point P, tangents PA and PB touch a circle with centre O. If ∠APB = 70°, find ∠AOB.

Answers

90°
9.5 cm
25 cm
13 cm
5 cm
110°

⚠️ Common mistakes

👉 Wrong: A tangent is perpendicular to every radius.
Correct: A tangent is perpendicular only to the radius through the point of contact.

👉 Wrong: A tangent can be drawn from a point inside a circle.
Correct: No real tangent can be drawn from a point inside a circle. One tangent can be drawn at a point on the circle, and two tangents can be drawn from a point outside the circle.

👉 Wrong: Two tangents from the same external point may have different lengths.
Correct: They are always equal in length.

👉 Wrong: In Theorem 41, OA = OB directly gives PA = PB.
Correct: First prove the two right triangles congruent by RHS, and then use corresponding sides.

👉 Wrong: When two circles touch, the centres and point of contact must always occur in the same order.
Correct: In external contact the point of contact lies between the centres. In internal contact the centres lie on the same side of the point of contact. In both cases, all three points are collinear.

📚 Important points to remember

👉 The radius through the point of contact is perpendicular to the tangent.

👉 A line perpendicular to a radius at its endpoint on the circle is a tangent.

👉 Only one tangent can be drawn at a fixed point on a circle.

👉 Two tangents can be drawn from an external point to a circle.

👉 Tangents drawn from the same external point are equal: PA = PB.

👉 The line joining the external point to the centre bisects the angle between the two radii to the points of contact and the angle between the two tangents.

👉 If two circles touch, their centres and point of contact are collinear.

👉 For external contact, distance between centres = sum of radii.

👉 For internal contact, distance between centres = difference of radii.