Tangents from an External Point
When a point lies outside a circle, exactly two tangents can be drawn from that point to the circle. These tangents have several important properties that are frequently tested in NCERT Class 10 board exams.
The most important theorem states that the lengths of two tangents drawn from an external point to a circle are equal. This result leads to several useful corollaries about angles, line segments, and quadrilateral properties.
This topic builds on the concept of a tangent to a circle (a line that touches the circle at exactly one point) and the theorem that a tangent is perpendicular to the radius at the point of contact.
What is Tangents from an External Point?
Definition: A tangent from an external point is a line drawn from a point outside a circle that touches the circle at exactly one point (the point of tangency).
Key Facts:
- From an external point, exactly two tangents can be drawn to a circle.
- From a point on the circle, exactly one tangent can be drawn.
- From a point inside the circle, no tangent can be drawn.
Key Terms:
- External Point (P) — the point outside the circle from which tangents are drawn
- Point of Tangency (A, B) — the points where the tangents touch the circle
- Centre (O) — the centre of the circle
- Radius (r) — OA = OB = radius
- Tangent Length — the distance from the external point to the point of tangency (PA or PB)
Tangents from an External Point Formula
Theorem 1: Equal Tangent Lengths
PA = PB
The lengths of two tangents from an external point to a circle are equal.
Theorem 2: Angle Properties
Angle OAP = Angle OBP = 90 deg
The tangent is perpendicular to the radius at the point of contact.
Corollary 1:
OP bisects angle APB and angle AOB
The line joining the centre O and the external point P bisects the angle between the two tangents and the angle between the two radii.
Corollary 2:
- Angle APB + Angle AOB = 180 deg
- In quadrilateral OAPB: angle OAP + angle OBP + angle APB + angle AOB = 360 deg
- Since angle OAP = angle OBP = 90 deg: angle APB + angle AOB = 180 deg
Length of Tangent:
PA = sqrt(OP squared minus r squared)
(By Pythagoras theorem in triangle OAP)
Derivation and Proof
Proof: Tangent lengths from an external point are equal (PA = PB)
- Given: A circle with centre O. P is an external point. PA and PB are tangents to the circle at points A and B respectively.
- To Prove: PA = PB
- Construction: Join OA, OB, and OP.
- Proof:
- In triangles OAP and OBP:
- OA = OB (radii of the same circle)
- Angle OAP = Angle OBP = 90 deg (tangent is perpendicular to radius at point of contact)
- OP = OP (common side)
- By RHS congruence: Triangle OAP is congruent to Triangle OBP
- Therefore, PA = PB (corresponding parts of congruent triangles)
Hence proved: PA = PB.
Additional Results from the Congruence:
- Angle OPA = Angle OPB → OP bisects angle APB
- Angle AOP = Angle BOP → OP bisects angle AOB
Types and Properties
Common Problem Types on Tangents from External Points:
| Type | What is Given | What to Find |
|---|---|---|
| Finding tangent length | Distance OP and radius r | PA = sqrt(OP squared minus r squared) |
| Finding angle between tangents | Angle AOB or radius and OP | Angle APB = 180 minus angle AOB |
| Tangent to incircle/circumscribed circle | Triangle with incircle | Tangent lengths from each vertex |
| Tangent to circumscribed quadrilateral | Quadrilateral around a circle | Sum of opposite sides are equal |
| Proving tangent properties | Circle with tangent and secant | Prove equal lengths, angles, etc. |
Tangent Property in Polygons:
- Triangle with incircle: If a circle is inscribed in a triangle with sides a, b, c, the tangent lengths from each vertex can be found using the property that tangent lengths from the same external point are equal.
- Quadrilateral circumscribing a circle: If a quadrilateral ABCD circumscribes a circle, then AB + CD = BC + DA (sum of opposite sides are equal).
Solved Examples
Example 1: Finding Tangent Length
Problem: A circle has centre O and radius 5 cm. A point P is 13 cm from the centre. Find the length of the tangent from P to the circle.
Solution:
Given: r = 5 cm, OP = 13 cm
Step 1: In right triangle OAP (angle OAP = 90 deg):
Step 2: PA squared = OP squared minus OA squared = 169 minus 25 = 144
Step 3: PA = sqrt(144) = 12 cm
Answer: Length of tangent = 12 cm
Example 2: Proving Equal Tangent Lengths
Problem: Prove that the tangents drawn from an external point to a circle are equal.
Solution:
Given: Circle with centre O. PA and PB are tangents from external point P touching the circle at A and B.
To Prove: PA = PB
Proof:
- Join OA, OB, OP.
- In triangles OAP and OBP:
- OA = OB (radii)
- Angle OAP = Angle OBP = 90 deg (tangent perpendicular to radius)
- OP = OP (common)
- By RHS congruence: Triangle OAP congruent to Triangle OBP
- Therefore PA = PB (CPCT)
Hence proved.
Example 3: Finding Angle Between Tangents
Problem: Two tangents PA and PB are drawn from an external point P to a circle with centre O. If angle AOB = 120 deg, find angle APB.
Solution:
Given: angle AOB = 120 deg
Step 1: In quadrilateral OAPB: angle OAP + angle OBP + angle APB + angle AOB = 360 deg
Step 2: 90 + 90 + angle APB + 120 = 360
Step 3: angle APB = 360 minus 300 = 60 deg
Answer: Angle between tangents = 60 degrees
Example 4: Tangents from a Point — Finding Radius
Problem: From a point P, the length of tangent to a circle is 24 cm and the distance from P to the centre is 25 cm. Find the radius.
Solution:
Given: PA = 24 cm, OP = 25 cm
Step 1: In right triangle OAP: OA squared = OP squared minus PA squared
Step 2: OA squared = 625 minus 576 = 49
Step 3: OA = sqrt(49) = 7 cm
Answer: Radius = 7 cm
Example 5: Incircle of a Triangle
Problem: A circle is inscribed in a triangle ABC with AB = 10 cm, BC = 8 cm, and CA = 12 cm. Find the lengths of tangent segments from each vertex.
Solution:
Given: Let the circle touch AB at P, BC at Q, and CA at R.
Step 1: Let AP = AR = x, BP = BQ = y, CQ = CR = z (tangent lengths from same external point are equal).
Step 2: From the sides:
- AB: x + y = 10 ... (i)
- BC: y + z = 8 ... (ii)
- CA: z + x = 12 ... (iii)
Step 3: Add all three: 2(x + y + z) = 30 → x + y + z = 15
Step 4: From (i): z = 15 minus 10 = 5 cm
Step 5: From (ii): x = 15 minus 8 = 7 cm
Step 6: From (iii): y = 15 minus 12 = 3 cm
Answer: AP = AR = 7 cm, BP = BQ = 3 cm, CQ = CR = 5 cm
Example 6: Circumscribed Quadrilateral
Problem: A quadrilateral ABCD circumscribes a circle. If AB = 6 cm, CD = 8 cm, and BC = 7 cm, find AD.
Solution:
Given: ABCD circumscribes a circle.
Property: AB + CD = BC + AD (sum of opposite sides of a circumscribed quadrilateral are equal).
Step 1: 6 + 8 = 7 + AD
Step 2: 14 = 7 + AD → AD = 7 cm
Answer: AD = 7 cm
Example 7: Angle When Tangents Are Perpendicular
Problem: Two tangents PA and PB are drawn to a circle with centre O such that angle APB = 90 degrees. Find angle AOB.
Solution:
Given: angle APB = 90 deg
Step 1: angle AOB + angle APB = 180 deg (proved earlier)
Step 2: angle AOB = 180 minus 90 = 90 deg
Answer: angle AOB = 90 degrees
Observation: When the tangents are perpendicular to each other, the angle at the centre between the radii is also 90 degrees.
Example 8: OP as Diameter of Circle Through O, A, P, B
Problem: Two tangents PA and PB are drawn to a circle with centre O, radius 3 cm, and OP = 5 cm. Find the area of the quadrilateral OAPB.
Solution:
Given: r = 3, OP = 5
Step 1: PA = sqrt(25 minus 9) = sqrt(16) = 4 cm
Step 2: Quadrilateral OAPB consists of two right triangles: OAP and OBP.
Step 3: Area of triangle OAP = (1/2) x OA x PA = (1/2) x 3 x 4 = 6 sq cm
Step 4: Area of triangle OBP = 6 sq cm (congruent to triangle OAP)
Step 5: Total area = 6 + 6 = 12 sq cm
Answer: Area of quadrilateral OAPB = 12 cm squared
Example 9: Tangent Property — Proof Based
Problem: Prove that the angle between two tangents drawn from an external point is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution:
Given: PA and PB are tangents from P. A and B are points of contact. O is the centre.
To Prove: angle APB + angle AOB = 180 deg
Proof:
- In quadrilateral OAPB: sum of all angles = 360 deg
- angle OAP + angle OBP + angle APB + angle AOB = 360 deg
- angle OAP = 90 deg (tangent perpendicular to radius)
- angle OBP = 90 deg
- 90 + 90 + angle APB + angle AOB = 360
- angle APB + angle AOB = 180 deg
Hence proved.
Example 10: NCERT — Parallelogram With Tangents
Problem: If a parallelogram ABCD circumscribes a circle, prove that it is a rhombus.
Solution:
Given: ABCD is a parallelogram circumscribing a circle.
To Prove: ABCD is a rhombus (all sides equal).
Proof:
- Since ABCD circumscribes a circle: AB + CD = BC + DA ... (property of circumscribed quadrilateral)
- Since ABCD is a parallelogram: AB = CD and BC = DA
- Substituting: AB + AB = BC + BC → 2AB = 2BC → AB = BC
- Therefore: AB = BC = CD = DA
- A parallelogram with all sides equal is a rhombus.
Hence proved.
Real-World Applications
Real-life applications:
- Gear Design: Tangent lines determine how gears mesh — the point of tangency is where force is transmitted between gears.
- Optics: Light rays tangent to curved mirrors and lenses help determine reflection and refraction angles.
- Road Design: When a road runs tangent to a roundabout, the tangent length determines the entry/exit geometry.
- Engineering: Belt drives around pulleys follow tangent paths. The tangent length determines belt length.
- Architecture: Arched doorways and bridges use tangent properties for structural calculations.
Key Points to Remember
- From an external point, exactly two tangents can be drawn to a circle.
- PA = PB — tangent lengths from the same external point are equal.
- Tangent is perpendicular to the radius at the point of contact: angle OAP = 90 deg.
- angle APB + angle AOB = 180 deg — the angle between tangents and angle at centre are supplementary.
- OP bisects both angle APB and angle AOB.
- Length of tangent: PA = sqrt(OP squared minus r squared).
- For a triangle with incircle: tangent lengths from each vertex satisfy x + y = a, y + z = b, z + x = c.
- For a quadrilateral circumscribing a circle: AB + CD = BC + DA.
- If a parallelogram circumscribes a circle, it must be a rhombus.
- These theorems are frequently asked as 3-mark and 5-mark proof questions in board exams.
Practice Problems
- A point P is 17 cm from the centre of a circle of radius 8 cm. Find the length of the tangent from P.
- Two tangents PA and PB are drawn from point P to a circle with centre O. If angle APB = 50 deg, find angle AOB.
- A circle is inscribed in triangle PQR with PQ = 7 cm, QR = 9 cm, PR = 11 cm. Find the tangent lengths from each vertex.
- If a quadrilateral PQRS circumscribes a circle with PQ = 5, QR = 6, RS = 8, find SP.
- Prove that a parallelogram circumscribing a circle is a rhombus.
- 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.
- If OP = 10 cm and r = 6 cm, find the area of the quadrilateral OAPB formed by the two tangents and two radii.
Frequently Asked Questions
Q1. How many tangents can be drawn from an external point?
Exactly two tangents can be drawn from an external point to a circle.
Q2. Are the two tangent lengths from an external point always equal?
Yes. This is a theorem: PA = PB, where PA and PB are tangents from external point P to the circle touching it at A and B. The proof uses RHS congruence of triangles OAP and OBP.
Q3. What is the angle between the tangent and the radius at the point of contact?
The tangent is always perpendicular to the radius at the point of contact. So the angle = 90 degrees.
Q4. What is the relationship between angle APB and angle AOB?
They are supplementary: angle APB + angle AOB = 180 degrees. This follows from the angle sum property of the quadrilateral OAPB.
Q5. How do you find the length of a tangent from a point to a circle?
Use Pythagoras theorem: PA = sqrt(OP squared minus r squared), where OP is the distance from the point to the centre and r is the radius.
Q6. What is the property of a quadrilateral circumscribing a circle?
If a quadrilateral ABCD circumscribes a circle, then the sum of opposite sides are equal: AB + CD = BC + DA.
Q7. What is an incircle?
An incircle is a circle inscribed inside a polygon (usually a triangle) such that it touches all sides of the polygon. Each side of the polygon is a tangent to the incircle.
Q8. If a parallelogram circumscribes a circle, what shape must it be?
A rhombus. Since AB + CD = BC + DA and AB = CD, BC = DA (parallelogram properties), we get 2AB = 2BC, so AB = BC. All sides are equal, making it a rhombus.
Q9. Can a rectangle circumscribe a circle?
Yes, but only if the rectangle is a square. For a rectangle circumscribing a circle: AB + CD = BC + DA. Since AB = CD and BC = DA: 2AB = 2BC, so AB = BC. A rectangle with all sides equal is a square.
Q10. How is this theorem used in construction problems?
To construct tangents from an external point: (1) Join OP, (2) Find the midpoint M of OP, (3) Draw a circle with centre M and radius MO, (4) This circle intersects the given circle at points A and B, (5) PA and PB are the required tangents.
Related Topics
- Tangent to a Circle
- Tangent is Perpendicular to Radius
- 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
