Tangent is Perpendicular to Radius
One of the most important results in Class 10 Circles is that a tangent to a circle is perpendicular to the radius at the point of contact. This theorem is the foundation for almost all tangent-related problems in CBSE board exams.
This property connects two geometric concepts — the tangent line that touches the circle at exactly one point, and the radius drawn to that same point. The perpendicularity between them leads to right-angled triangles, enabling the use of Pythagoras theorem and trigonometric ratios.
This result is used to prove other theorems, including the equal tangent lengths theorem and properties of tangents from an external point.
What is Tangent Perpendicular to Radius - Theorem, Proof & Solved Examples?
Theorem (NCERT Theorem 10.1): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
In symbols: If O is the centre, T is the point of contact, and PT is the tangent, then:
OT ⊥ PT (i.e., ∠OTP = 90°)
Converse: A line drawn through the end of a radius and perpendicular to it is a tangent to the circle.
Key observations:
- The radius OT is the shortest distance from centre O to the tangent line PT.
- This means OT is the perpendicular distance from O to the line PT.
- Any other line from O to a point on PT would be longer than OT (oblique distance > perpendicular distance).
Tangent is Perpendicular to Radius Formula
Key Results from This Theorem:
∠OTP = 90° (radius ⊥ tangent at point of contact)
Consequences:
- In triangle OTP: OT² + PT² = OP² (by Pythagoras theorem)
- If OT = r (radius) and OP = d (distance from centre to external point), then: PT = √(d² − r²)
Tangent length from external point: PT = √(OP² − OT²) = √(d² − r²)
Where:
- PT = length of tangent from P to point of contact T
- OP = d = distance from external point P to centre O
- OT = r = radius of the circle
Derivation and Proof
Proof: Tangent at any point is perpendicular to the radius (Proof by Contradiction)
Given: A circle with centre O, radius r. PT is a tangent at point T.
To prove: OT ⊥ PT
- Assume OT is NOT perpendicular to PT.
- Then there exists another point Q on the tangent line PT such that OQ ⊥ PT.
- In right triangle OQT (right angle at Q): the hypotenuse OT is longer than leg OQ.
- So OQ < OT, i.e., OQ < r.
- This means Q lies inside the circle (since its distance from O is less than r).
- But Q lies on the tangent PT. If Q is inside the circle, then the line PT must intersect the circle at two points (entering and exiting the circle).
- This contradicts the definition of a tangent, which touches the circle at only one point.
- Therefore, our assumption is wrong.
- Hence, OT ⊥ PT.
Proof of the Converse:
- Given: A line l passes through point T on the circle and is perpendicular to radius OT.
- Take any point Q on line l, where Q ≠ T.
- In right triangle OQT: OQ > OT (hypotenuse > leg).
- So OQ > r, meaning Q lies outside the circle.
- Every point on line l (except T) lies outside the circle.
- Therefore, line l meets the circle at exactly one point T.
- Hence, l is a tangent to the circle at T.
Types and Properties
Applications of Tangent-Radius Perpendicularity:
- Finding tangent length: Use Pythagoras theorem in △OTP to find PT when OP and OT are known.
- Proving equal tangent lengths: This theorem is used in the proof that tangents from an external point are equal.
- Finding angles: Since ∠OTP = 90°, we can find other angles in the figure using angle sum property.
- Construction problems: Draw a tangent at a point by first drawing the radius and then constructing a perpendicular.
Summary of angle relationships:
| Configuration | Angle |
|---|---|
| Radius and tangent at point of contact | 90° |
| Two tangents from external point (between them) | Depends on OP and r |
| Angle at centre (∠AOB) + angle between tangents (∠APB) | 180° (supplementary) |
Solved Examples
Example 1: Finding Tangent Length
Problem: A point P is 13 cm from the centre O of a circle of radius 5 cm. Find the length of the tangent from P to the circle.
Solution:
Given: OP = 13 cm, OT = 5 cm (radius), ∠OTP = 90°
Using Pythagoras theorem in △OTP:
- PT² = OP² − OT²
- PT² = 13² − 5²
- PT² = 169 − 25 = 144
- PT = √144 = 12 cm
Answer: Length of tangent = 12 cm
Example 2: Finding Distance from Centre
Problem: The length of a tangent from a point to a circle of radius 7 cm is 24 cm. Find the distance of the point from the centre.
Solution:
Given: OT = 7 cm, PT = 24 cm, ∠OTP = 90°
Using Pythagoras theorem:
- OP² = OT² + PT²
- OP² = 7² + 24²
- OP² = 49 + 576 = 625
- OP = √625 = 25 cm
Answer: Distance from centre = 25 cm
Example 3: Checking if a Line is Tangent
Problem: A circle has centre O and radius 6 cm. A point T lies on the circle, and PT = 8 cm. If OP = 10 cm, prove that PT is a tangent.
Solution:
Check if ∠OTP = 90°:
- OT² + PT² = 6² + 8² = 36 + 64 = 100
- OP² = 10² = 100
- Since OT² + PT² = OP², triangle OTP is right-angled at T.
Therefore: OT ⊥ PT, and since T is on the circle, PT is a tangent by the converse theorem.
Answer: PT is a tangent to the circle at T.
Example 4: Finding Radius from Tangent and External Distance
Problem: From an external point P, 17 cm from the centre, a tangent of length 15 cm is drawn. Find the radius of the circle.
Solution:
Given: OP = 17 cm, PT = 15 cm
Using Pythagoras theorem:
- OT² = OP² − PT²
- OT² = 17² − 15²
- OT² = 289 − 225 = 64
- OT = √64 = 8 cm
Answer: Radius = 8 cm
Example 5: Angle Between Tangent and Chord
Problem: PQ is a tangent to a circle with centre O at point Q. If OQ = 5 cm and ∠OPQ = 30°, find PQ and OP.
Solution:
Given: OQ = 5 cm, ∠OPQ = 30°, ∠OQP = 90° (tangent ⊥ radius)
In right △OQP:
- tan 30° = OQ/PQ → 1/√3 = 5/PQ → PQ = 5√3 cm
- sin 30° = OQ/OP → 1/2 = 5/OP → OP = 10 cm
Answer: PQ = 5√3 cm ≈ 8.66 cm, OP = 10 cm
Example 6: Two Tangents and Angle at Centre
Problem: Two tangents PA and PB are drawn from an external point P to a circle with centre O. If ∠APB = 60°, find ∠AOB.
Solution:
Given: ∠APB = 60°
Since OA ⊥ PA and OB ⊥ PB:
- ∠OAP = 90° and ∠OBP = 90°
In quadrilateral OAPB:
- Sum of angles = 360°
- ∠OAP + ∠APB + ∠OBP + ∠AOB = 360°
- 90° + 60° + 90° + ∠AOB = 360°
- ∠AOB = 360° − 240° = 120°
Answer: ∠AOB = 120°
Example 7: Tangent from External Point — Full Solution
Problem: From an external point P, two tangents PA and PB are drawn to a circle of radius 4 cm with centre O. If OP = 8 cm, find PA, ∠APB, and the area of △OAP.
Solution:
(i) Finding PA:
- PA² = OP² − OA² = 64 − 16 = 48
- PA = √48 = 4√3 cm
(ii) Finding ∠APB:
- cos(∠OPA) = PA/OP = 4√3/8 = √3/2
- ∠OPA = 30°
- ∠APB = 2 × ∠OPA = 60° (since OP bisects ∠APB)
(iii) Area of △OAP:
- Area = ½ × OA × PA = ½ × 4 × 4√3 = 8√3 cm²
Answer: PA = 4√3 cm, ∠APB = 60°, Area = 8√3 cm²
Example 8: Perpendicularity in Inscribed Quadrilateral
Problem: A circle is inscribed in a triangle ABC, touching AB at P, BC at Q, and CA at R. If ∠A = 50°, find ∠BOC where O is the centre of the inscribed circle.
Solution:
Since OP ⊥ AB and OR ⊥ AC: (tangent ⊥ radius)
- In quadrilateral APOR: ∠OPA = 90°, ∠ORA = 90°, ∠A = 50°
- ∠POR = 360° − 90° − 90° − 50° = 130°
The angle ∠BOC at centre:
- ∠BOC = 180° − ∠A/2 ... (standard result for incircle)
- Actually, using the quadrilateral BQOC: ∠OQB = 90°, ∠OCQ = 90°
- ∠B + ∠C = 180° − 50° = 130°
- In quad BQOC: ∠QOC + ∠OQB + ∠B' + ∠OCQ = 360° (where ∠B' is part of ∠B)
Using the direct result: ∠BOC = 90° + ∠A/2 = 90° + 25° = 115°
Answer: ∠BOC = 115°
Real-World Applications
Applications of the tangent-radius perpendicularity theorem:
- Engineering: Wheels on tracks — the contact point between a wheel and a flat surface forms a tangent, and the spoke (radius) is perpendicular to the surface.
- Constructions: Drawing accurate tangent lines to circles in architectural and mechanical drawings.
- Proving other theorems: This is the basis for proving equal tangent lengths and angle relationships in circles.
- Gear design: The tangent-radius relationship determines how gears mesh and transfer motion.
- Optics: Light reflecting off a curved mirror obeys this perpendicularity — the normal to a curved surface at any point is along the radius.
Key Points to Remember
- A tangent at any point of a circle is perpendicular to the radius through the point of contact.
- Converse: A line perpendicular to the radius at its endpoint on the circle is a tangent.
- The proof uses contradiction: assuming non-perpendicularity leads to the tangent intersecting the circle at two points.
- In right triangle OTP: PT² = OP² − OT² (tangent length formula).
- ∠OTP = 90° creates a right angle that is used in numerous circle geometry problems.
- If two tangents are drawn from an external point P: ∠OAP = ∠OBP = 90°.
- In quadrilateral OAPB: ∠AOB + ∠APB = 180° (supplementary).
- The radius is the shortest distance from the centre to the tangent line.
- This theorem is NCERT Theorem 10.1 and is frequently asked in board exams (2–3 marks).
Practice Problems
- A point P is 26 cm from the centre of a circle of radius 10 cm. Find the length of the tangent from P.
- The tangent from an external point to a circle of radius 12 cm has length 16 cm. Find the distance of the point from the centre.
- Two tangents TP and TQ are drawn from T to a circle with centre O. If ∠PTQ = 70°, find ∠POQ.
- Prove that the tangent at any point of a circle is perpendicular to the radius at the point of contact.
- From a point P, 25 cm from the centre O of a circle of radius 7 cm, tangent PT is drawn. Find the area of triangle OPT.
- If OA = 5 cm, OP = 13 cm, and PA is a tangent, find sin(∠OPA) and cos(∠OPA).
Frequently Asked Questions
Q1. Why is the tangent perpendicular to the radius?
If the tangent were not perpendicular, a shorter line from the centre to the tangent would exist, meaning that line's foot would be inside the circle. This would make the tangent cross the circle at two points, contradicting the definition of a tangent.
Q2. Is this theorem always true for every circle?
Yes. This theorem holds for every circle regardless of size. At every point on any circle, the tangent is perpendicular to the radius drawn to that point.
Q3. How is this used to find tangent length?
Since ∠OTP = 90°, triangle OTP is right-angled. By Pythagoras theorem: PT = √(OP² − OT²), where OP is the distance from the external point to the centre and OT is the radius.
Q4. What is the converse of this theorem?
The converse states: If a line is perpendicular to a radius at its endpoint on the circle, then that line is a tangent to the circle. This is used in construction problems to draw tangents.
Q5. Can a tangent be drawn from a point inside the circle?
No. From a point inside the circle, no tangent can be drawn. From a point on the circle, exactly one tangent exists. From an external point, exactly two tangents can be drawn.
Q6. How does ∠AOB relate to ∠APB when PA and PB are tangents?
∠AOB + ∠APB = 180°. This follows because in quadrilateral OAPB, ∠OAP = ∠OBP = 90°, so ∠AOB + ∠APB = 360° − 180° = 180°.
Q7. Is this theorem in the CBSE board exam syllabus?
Yes. This is NCERT Theorem 10.1 (Chapter 10: Circles). Both the statement and proof are important for board exams. Problems based on this theorem carry 2-3 marks.
Q8. How is this theorem related to the equal tangent lengths theorem?
The equal tangent lengths proof uses this theorem as a step. In triangles OAP and OBP, ∠OAP = ∠OBP = 90° (from this theorem), which helps prove the triangles are congruent, giving PA = PB.
Related Topics
- Tangent to a Circle
- Tangents from an External Point
- Number of Tangents from a Point
- Circle Theorems Introduction
- Angle Subtended by a Chord
- Equal Chords and Equal Angles
- Perpendicular from Centre to Chord
- Angles in the Same Segment
- Angle in a Semicircle
- Cyclic Quadrilateral
- Tangent-Secant Relationship
- Angle Between Two Tangents
- Important Circle Theorem Problems
- Properties of Tangents - Summary
