Number of Tangents from a Point

Class 10Circles

The number of tangents that can be drawn to a circle from a point depends on the position of the point relative to the circle. This is a fundamental concept in Class 10 CBSE Chapter 10 (Circles).

There are exactly three cases: the point lies inside the circle, on the circle, or outside the circle. Each case yields a different number of tangents — 0, 1, or 2 respectively.

The case of two tangents from an external point leads to the important result that these tangents are equal in length, which is NCERT Theorem 10.2.

What is Number of Tangents from a Point - Cases, Theorems & Solved Examples?

Definition: A tangent to a circle is a line that touches the circle at exactly one point (the point of contact). The number of tangents from a given point depends on its position relative to the circle.

Three Cases:

Position of Point P	Condition	Number of Tangents
Inside the circle	OP < r	0 (no tangent)
On the circle	OP = r	1 (exactly one tangent)
Outside the circle	OP > r	2 (exactly two tangents)

Key Terms:

Secant — a line that intersects the circle at two points (passes through the circle)
Point of contact (tangency) — the single point where a tangent meets the circle
External point — a point outside the circle (OP > r)

Number of Tangents from a Point Formula

Theorem (NCERT Theorem 10.2): The lengths of tangents drawn from an external point to a circle are equal.

If PA and PB are tangents from external point P to a circle with centre O, then PA = PB

Additional properties when two tangents are drawn from P:

PA = PB (equal tangent lengths)
∠OPA = ∠OPB (OP bisects the angle between tangents)
∠OAP = ∠OBP = 90° (tangent ⊥ radius)
∠AOB + ∠APB = 180° (supplementary angles)
OP bisects AB perpendicularly

Tangent length formula:

PA = PB = √(OP² − r²)

Derivation and Proof

Proof: Tangents from an external point are equal (NCERT Theorem 10.2)

Given: PA and PB are tangents from external point P to a circle with centre O, touching at A and B.

To prove: PA = PB

Join OA, OB, and OP.
Since PA is a tangent and OA is a radius: ∠OAP = 90° (Theorem 10.1)
Since PB is a tangent and OB is a radius: ∠OBP = 90°
In right triangles △OAP and △OBP:
- OA = OB (radii of same circle)
- OP = OP (common side)
- ∠OAP = ∠OBP = 90°
By RHS congruence: △OAP ≅ △OBP
By CPCT: PA = PB

Additional results from congruence (CPCT):

∠OPA = ∠OPB → OP bisects ∠APB
∠AOP = ∠BOP → OP bisects ∠AOB

Why no tangent from an internal point:

Let P be inside the circle (OP < r).
Any line through P must enter and exit the circle (since P is inside).
Such a line intersects the circle at 2 points, making it a secant, not a tangent.
Therefore, no tangent can be drawn from an internal point.

Types and Properties

Summary of all cases:

Case 1: Point inside the circle (OP < r)

Every line through P intersects the circle at 2 points.
All such lines are secants.
Number of tangents = 0.

Case 2: Point on the circle (OP = r)

Exactly one tangent exists — perpendicular to the radius at that point.
Number of tangents = 1.

Case 3: Point outside the circle (OP > r)

Exactly two tangents can be drawn.
Both tangents have equal length.
The line OP bisects the angle between the two tangents.
Number of tangents = 2.

Solved Examples

Example 1: Determining Number of Tangents

Problem: A circle has centre O and radius 5 cm. Determine the number of tangents from point P if (a) OP = 3 cm, (b) OP = 5 cm, (c) OP = 8 cm.

Solution:

(a) OP = 3 cm < 5 cm (radius): P is inside the circle → 0 tangents

(b) OP = 5 cm = radius: P is on the circle → 1 tangent

(c) OP = 8 cm > 5 cm: P is outside the circle → 2 tangents

Example 2: Finding Equal Tangent Lengths

Problem: From an external point P, two tangents PA and PB are drawn to a circle of radius 6 cm with centre O. If OP = 10 cm, find PA and PB.

Solution:

Given: OA = 6 cm (radius), OP = 10 cm

Since ∠OAP = 90°:

PA² = OP² − OA² = 100 − 36 = 64
PA = 8 cm

Since tangents from external point are equal: PB = PA = 8 cm

Answer: PA = PB = 8 cm

Example 3: Finding Angle Between Two Tangents

Problem: Two tangents PA and PB are drawn from P to a circle with centre O. If ∠AOB = 110°, find ∠APB.

Solution:

Using: ∠AOB + ∠APB = 180° (supplementary in quadrilateral OAPB)

110° + ∠APB = 180°
∠APB = 180° − 110° = 70°

Answer: ∠APB = 70°

Example 4: Tangents to Two Concentric Circles

Problem: Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that is tangent to the smaller circle.

Solution:

Given: R = 5 cm (larger), r = 3 cm (smaller), centre O

Let AB be a chord of the larger circle, tangent to the smaller circle at T.

OT ⊥ AB (tangent ⊥ radius of smaller circle)
OT = 3 cm
OA = 5 cm (radius of larger circle)

In right triangle OTA:

AT² = OA² − OT² = 25 − 9 = 16
AT = 4 cm

Since OT ⊥ AB, T is the midpoint of AB:

AB = 2 × AT = 2 × 4 = 8 cm

Answer: Length of chord = 8 cm

Example 5: Circle Inscribed in a Triangle — Tangent Lengths

Problem: A circle is inscribed in triangle ABC, touching AB at P, BC at Q, and CA at R. If AB = 8 cm, BC = 10 cm, and CA = 12 cm, find AP, BQ, and CR.

Solution:

Using equal tangent lengths:

AP = AR = x (tangents from A)
BP = BQ = y (tangents from B)
CQ = CR = z (tangents from C)

Setting up equations:

x + y = AB = 8 ... (1)
y + z = BC = 10 ... (2)
z + x = CA = 12 ... (3)

Adding all three: 2(x + y + z) = 30 → x + y + z = 15

Solving:

z = 15 − 8 = 7 (from equation 1)
x = 15 − 10 = 5 (from equation 2)
y = 15 − 12 = 3 (from equation 3)

Answer: AP = 5 cm, BQ = 3 cm, CR = 7 cm

Example 6: Proving PA = PB Using Theorem

Problem: Prove that the tangents drawn from an external point to a circle are equal.

Solution (Proof):

Given: PA, PB are tangents from P to circle with centre O.

To prove: PA = PB

Join OA, OB, OP.
∠OAP = 90° and ∠OBP = 90° (tangent ⊥ radius)
In △OAP and △OBP:
- OA = OB (radii)
- OP = OP (common)
- ∠OAP = ∠OBP = 90°
△OAP ≅ △OBP (RHS congruence)
PA = PB (CPCT)

Hence proved.

Example 7: Quadrilateral with Tangent Sides

Problem: ABCD is a quadrilateral circumscribing a circle. Prove that AB + CD = AD + BC.

Solution:

Let the circle touch AB at P, BC at Q, CD at R, DA at S.

Using equal tangent lengths:

AP = AS (tangents from A)
BP = BQ (tangents from B)
CR = CQ (tangents from C)
DR = DS (tangents from D)

Now:

AB + CD = (AP + BP) + (CR + DR)
= AS + BQ + CQ + DS
= (AS + DS) + (BQ + CQ)
= AD + BC

Hence: AB + CD = AD + BC. Proved.

Example 8: Finding Angle with Equal Tangents

Problem: PA and PB are tangents from P to a circle with centre O, such that ∠APB = 80°. Find ∠OAB.

Solution:

Step 1: ∠AOB = 180° − ∠APB = 180° − 80° = 100°

Step 2: In triangle OAB, OA = OB (radii), so △OAB is isosceles.

∠OAB = ∠OBA = (180° − 100°)/2 = 80°/2 = 40°

Answer: ∠OAB = 40°

Real-World Applications

Applications of tangent properties:

Wheel mechanics: A wheel touching a flat road forms a tangent at the contact point — the force is perpendicular to the surface (along the radius).
Satellite orbits: A satellite's velocity at any point is tangential to its orbit.
Construction: Drawing common tangents to two circles (internal and external) is fundamental in gear and pulley design.
Lens design: The tangent at any point on a curved lens surface determines the angle of refraction.
Security cameras: The field of view from a camera mounted at an external point forms two tangent lines to a circular area.

Key Points to Remember

From a point inside a circle: 0 tangents.
From a point on a circle: 1 tangent (perpendicular to the radius).
From a point outside a circle: 2 tangents (equal in length).
Equal tangent lengths: PA = PB = √(OP² − r²).
∠OAP = ∠OBP = 90° (tangent ⊥ radius).
∠AOB + ∠APB = 180° in quadrilateral OAPB.
OP bisects ∠APB and ∠AOB.
For a circumscribed quadrilateral: AB + CD = AD + BC.
The proof uses RHS congruence of triangles OAP and OBP.
This theorem is NCERT Theorem 10.2 — a frequently asked board exam question.

Practice Problems

A circle has radius 7 cm. Determine the number of tangents from a point at distance (a) 5 cm, (b) 7 cm, (c) 12 cm from the centre.
From an external point 15 cm from the centre of a circle of radius 9 cm, find the length of each tangent.
Two tangents from P to a circle with centre O make an angle of 50° at P. Find ∠AOB.
A circle inscribed in a triangle touches the sides of lengths 6 cm, 8 cm, and 10 cm. Find the tangent lengths from each vertex.
Prove that a quadrilateral circumscribing a circle has AB + CD = AD + BC.
PA and PB are tangents from P. If PA = 12 cm and OP = 13 cm, find the radius of the circle.

Frequently Asked Questions

Q1. How many tangents can be drawn from a point inside a circle?

Zero. Any line through an internal point must intersect the circle at two points, making it a secant, not a tangent.

Q2. Why are tangents from an external point equal?

The proof uses RHS congruence: triangles OAP and OBP have OA = OB (radii), OP = OP (common), and ∠OAP = ∠OBP = 90°. So the triangles are congruent and PA = PB.

Q3. What is the relationship between ∠AOB and ∠APB?

They are supplementary: ∠AOB + ∠APB = 180°. This comes from the angle sum of quadrilateral OAPB where ∠OAP = ∠OBP = 90°.

Q4. Can three tangents be drawn from any point to a circle?

No. The maximum number of tangents from any point to a single circle is 2 (from an external point). From a point on the circle, only 1, and from inside, 0.

Q5. What is the tangent length formula?

If P is an external point at distance d from centre O, and r is the radius, then tangent length = √(d² − r²). This follows from Pythagoras theorem in the right triangle formed by the radius, tangent, and line joining P to O.

Q6. What is a common tangent to two circles?

A common tangent is a line tangent to both circles simultaneously. External common tangents do not pass between the circles; internal common tangents pass between them. Two circles can have 0, 1, 2, 3, or 4 common tangents depending on their relative positions.

Q7. How is this topic tested in board exams?

Common questions include: proving PA = PB, finding tangent lengths, proving AB + CD = AD + BC for circumscribed quadrilaterals, and finding angles in two-tangent configurations. These typically carry 3-5 marks.

Q8. What is the difference between a secant and a tangent?

A secant intersects the circle at 2 points. A tangent touches the circle at exactly 1 point. As a secant rotates about an external point, it approaches a tangent as the two intersection points merge into one.

Number of Tangents from a Point

What is Number of Tangents from a Point - Cases, Theorems & Solved Examples?

Number of Tangents from a Point Formula

Derivation and Proof

Types and Properties

Solved Examples

Example 1: Determining Number of Tangents

Example 2: Finding Equal Tangent Lengths

Example 3: Finding Angle Between Two Tangents

Example 4: Tangents to Two Concentric Circles

Example 5: Circle Inscribed in a Triangle — Tangent Lengths

Example 6: Proving PA = PB Using Theorem

Example 7: Quadrilateral with Tangent Sides

Example 8: Finding Angle with Equal Tangents

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. How many tangents can be drawn from a point inside a circle?

Q2. Why are tangents from an external point equal?

Q3. What is the relationship between ∠AOB and ∠APB?

Q4. Can three tangents be drawn from any point to a circle?

Q5. What is the tangent length formula?

Q6. What is a common tangent to two circles?

Q7. How is this topic tested in board exams?

Q8. What is the difference between a secant and a tangent?

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