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Constructing Tangents to a Circle

Class 10Constructions

A tangent to a circle is a line that touches the circle at exactly one point. In Class 10 Constructions, two types of tangent constructions are studied:

  • Tangent to a circle from a point on the circle.
  • Tangent to a circle from a point outside the circle (external point).

The construction from an external point produces two tangents of equal length. This construction uses the property that a tangent is perpendicular to the radius at the point of contact, and the angle in a semicircle is 90°.


These constructions are frequently asked in CBSE board exams and carry 4–5 marks.



The tangent construction from an external point is one of the most elegant constructions in geometry. It combines three fundamental properties: (1) the tangent is perpendicular to the radius, (2) the angle in a semicircle is 90 degrees, and (3) the perpendicular bisector construction for finding a midpoint.


Understanding the justification of each construction step is as important as performing the construction itself. CBSE board exams often ask for both the construction and the proof of why it works.

What is Constructing Tangents to a Circle?

Definition: A tangent to a circle is a straight line that touches the circle at exactly one point, called the point of contact or point of tangency.


Key properties used in construction:

  • The tangent at any point is perpendicular to the radius through that point.
  • From an external point, exactly two tangents can be drawn to a circle.
  • These two tangents are equal in length.
  • The angle in a semicircle is 90° — this property helps locate the points of tangency.

Constructing Tangents to a Circle Formula

Length of Tangent from External Point:

Length of tangent = √(d² − r²)


Where:

  • d = distance from the external point to the centre of the circle
  • r = radius of the circle

Why this formula works:

  • The tangent, radius, and line from centre to external point form a right triangle.
  • The right angle is at the point of tangency (tangent ⊥ radius).
  • By Pythagoras: d² = r² + (tangent length)².

Derivation and Proof

Justification — Construction from External Point:


Given circle with centre O and radius r, and external point P at distance d from O.

  1. Join OP.
  2. Find midpoint M of OP (by perpendicular bisector construction).
  3. With M as centre and MO as radius, draw a circle (this circle has diameter OP).
  4. This new circle intersects the given circle at points T₁ and T₂.
  5. Why T₁ and T₂ are points of tangency:
  6. T₁ lies on the circle with diameter OP, so ∠OT₁P = 90° (angle in semicircle).
  7. This means OT₁ ⊥ PT₁, i.e., the radius OT₁ is perpendicular to PT₁.
  8. Since PT₁ is perpendicular to the radius at the point on the circle, PT₁ is a tangent.
  9. Similarly, PT₂ is a tangent.
  10. Both tangents have equal length: PT₁ = PT₂ = √(d² − r²).


Why two tangent points exist:

  • The auxiliary circle (diameter OP) has radius OP/2, which is greater than r (since P is external, OP > r, so OP/2 > r/2).
  • The auxiliary circle passes through O (inside the original circle) and P (outside the original circle).
  • Since it crosses from inside to outside, it must intersect the original circle at exactly two points.
  • These two intersection points are the tangent points T₁ and T₂.

Types and Properties

Type 1: Tangent from a Point ON the Circle

  • The point of tangency is already known.
  • Construct the radius to that point, then draw a perpendicular to the radius at that point.
  • The perpendicular line is the tangent.
  • There is only one tangent at a given point on the circle.

Type 2: Tangent from an External Point

  • The external point P is at a distance greater than the radius from the centre.
  • Construct the perpendicular bisector of OP to find midpoint M.
  • Draw a circle with centre M and radius MO.
  • This circle intersects the original circle at two points — the points of tangency.
  • Two tangents are drawn, and they are equal in length.

PropertyPoint on CircleExternal Point
Number of tangents12
Key property usedTangent ⊥ radiusAngle in semicircle = 90°
Construction toolPerpendicular at a pointPerpendicular bisector + auxiliary circle


Number of tangents based on position of point:

  • Point inside the circle (d < r): Zero tangents. Every line through an internal point is a secant.
  • Point on the circle (d = r): Exactly one tangent, perpendicular to the radius.
  • Point outside the circle (d > r): Exactly two tangents, equal in length.

Angle between two tangents (from external point):

  • sin(half-angle) = r/d, so the full angle between tangents = 2 arcsin(r/d).
  • When d is very large compared to r, the angle approaches 0 degrees (tangents become nearly parallel).
  • When d is just slightly greater than r, the angle approaches 180 degrees (tangents open wide).

Methods

Construction A — Tangent from a Point on the Circle:

  1. Draw a circle with centre O and given radius.
  2. Mark the given point T on the circle.
  3. Join OT (this is the radius).
  4. At point T, construct a perpendicular to OT using compass (standard perpendicular construction at a point on a line).
  5. The perpendicular line through T is the required tangent.

Construction B — Two Tangents from an External Point:

  1. Draw a circle with centre O and radius r.
  2. Mark the external point P at distance d from O (d > r).
  3. Join OP.
  4. Find the midpoint M of OP using perpendicular bisector construction.
  5. With M as centre and MO (= MP = OP/2) as radius, draw a circle.
  6. This circle intersects the original circle at T₁ and T₂.
  7. Join PT₁ and PT₂.
  8. PT₁ and PT₂ are the two required tangents.

How to find perpendicular bisector (Step 4):

  1. With O as centre and radius more than half of OP, draw arcs on both sides.
  2. With P as centre and the same radius, draw arcs intersecting the previous ones.
  3. Join the two intersection points — this line bisects OP at M.


Verification Steps:

  1. After construction, measure PT₁ and PT₂ — they should be equal.
  2. Measure ∠OT₁P with a protractor — it should be 90°.
  3. Check: PT₁² + OT₁² should equal OP² (Pythagoras).

Common Mistakes:

  • Perpendicular bisector error: If M is not exactly the midpoint of OP, the auxiliary circle will not pass through the correct tangent points.
  • Wrong compass radius for auxiliary circle: The radius must be MO (= MP = OP/2), not OP or r.
  • Not drawing both tangent lines: The auxiliary circle intersects the original circle at two points — both must be joined to P.
  • Confusing radius and diameter: The auxiliary circle has OP as diameter (not radius).

Special Constructions:

  • Direct common tangent: For two circles, tangents that touch both circles on the same side.
  • Transverse common tangent: Tangents that cross between the two circles. These are advanced constructions beyond Class 10 scope.

Solved Examples

Example 1: Tangent at a Point on the Circle

Problem: Draw a circle of radius 4 cm and construct a tangent at a point T on the circle.


Solution:

Given:

  • Centre O, radius = 4 cm, point T on the circle

Steps:

  1. Draw circle with centre O and radius 4 cm.
  2. Mark any point T on the circumference.
  3. Join OT.
  4. At T, construct perpendicular to OT.

Result: The perpendicular at T is the tangent. It touches the circle only at T and is perpendicular to the radius OT.

Example 2: Two Tangents from External Point (d = 8 cm, r = 3 cm)

Problem: Draw a circle of radius 3 cm. From a point P at distance 8 cm from the centre, construct two tangents to the circle. Measure the tangent length.


Solution:

Given:

  • Radius r = 3 cm, OP = d = 8 cm

Steps:

  1. Draw circle with centre O, radius 3 cm.
  2. Mark point P such that OP = 8 cm.
  3. Join OP and find midpoint M (perpendicular bisector).
  4. With M as centre and MO = 4 cm as radius, draw auxiliary circle.
  5. Mark intersection points T₁ and T₂ with the original circle.
  6. Join PT₁ and PT₂.

Expected tangent length:

  • PT = √(d² − r²) = √(64 − 9) = √55 ≈ 7.42 cm

Example 3: Two Tangents (d = 6 cm, r = 3 cm)

Problem: Draw a circle of radius 3 cm. From a point 6 cm from the centre, construct two tangents.


Solution:

Given:

  • r = 3 cm, d = 6 cm

Steps:

  1. Draw circle with centre O, radius 3 cm.
  2. Mark P with OP = 6 cm. Join OP.
  3. Find midpoint M of OP. OM = MP = 3 cm.
  4. Draw auxiliary circle with centre M, radius 3 cm.
  5. This circle passes through O and P and intersects the original circle at T₁, T₂.
  6. Join PT₁, PT₂.

Tangent length: √(36 − 9) = √27 = 3√3 ≈ 5.2 cm

Angle between tangents: ∠T₁PT₂ = 2 × arcsin(r/d) = 2 × arcsin(1/2) = 2 × 30° = 60°.

Example 4: Tangent with Angle Calculation

Problem: A circle has radius 5 cm. From an external point P at distance 13 cm from the centre, tangents are drawn. Find the tangent length and the angle between the two tangents.


Solution:

Given:

  • r = 5 cm, d = 13 cm

Tangent length:

  • PT = √(13² − 5²) = √(169 − 25) = √144 = 12 cm

Angle calculation:

  • In right △OTP: tan(∠OPT) = OT/PT = 5/12
  • ∠OPT = arctan(5/12) ≈ 22.6°
  • ∠T₁PT₂ = 2 × ∠OPT ≈ 2 × 22.6° ≈ 45.2°

Note: The angle between tangents decreases as the external point moves farther away.

Example 5: NCERT Exercise 11.2 Q1 — Radius 6 cm, Point at 10 cm

Problem: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.


Solution:

Given:

  • r = 6 cm, d = 10 cm

Steps:

  1. Draw circle with centre O, radius 6 cm.
  2. Mark P with OP = 10 cm.
  3. Find midpoint M of OP. OM = 5 cm.
  4. Draw auxiliary circle: centre M, radius 5 cm.
  5. Auxiliary circle intersects original circle at T₁, T₂.
  6. Join PT₁ and PT₂.

Expected tangent length:

  • PT = √(100 − 36) = √64 = 8 cm

Verification: Measure PT₁ and PT₂ — both should be approximately 8 cm.

Example 6: Tangents from a Point Where d = 2r

Problem: A circle has radius 4 cm. Tangents are drawn from a point at distance 8 cm (= 2r) from the centre. Find tangent length and ∠T₁PT₂.


Solution:

Given:

  • r = 4 cm, d = 8 cm = 2r

Tangent length:

  • PT = √(64 − 16) = √48 = 4√3 ≈ 6.93 cm

Angle:

  • sin(∠OPT) = r/d = 4/8 = 1/2
  • ∠OPT = 30°
  • ∠T₁PT₂ = 2 × 30° = 60°

Observation: When d = 2r, the angle between tangents is always 60°.

Example 7: Tangents to Two Concentric Circles

Problem: Two concentric circles have radii 3 cm and 5 cm. Draw a tangent to the inner circle from a point on the outer circle.


Solution:

Given:

  • Inner circle: centre O, radius 3 cm
  • Outer circle: centre O, radius 5 cm
  • Point P is on the outer circle, so OP = 5 cm

Steps:

  1. Draw both concentric circles.
  2. Mark P on the outer circle.
  3. Find midpoint M of OP (OM = 2.5 cm).
  4. Draw auxiliary circle: centre M, radius 2.5 cm.
  5. This circle intersects the inner circle at T₁, T₂.
  6. Join PT₁ and PT₂.

Tangent length: √(25 − 9) = √16 = 4 cm

Example 8: Verifying Tangent is Perpendicular to Radius

Problem: After constructing tangents from P to a circle (centre O, radius 3 cm, OP = 5 cm), verify that ∠OT₁P = 90°.


Solution:

Measurements:

  • OT₁ = 3 cm (radius)
  • PT₁ = √(25 − 9) = 4 cm (tangent)
  • OP = 5 cm

Check Pythagoras:

  • OT₁² + PT₁² = 9 + 16 = 25 = OP²
  • Since OT₁² + PT₁² = OP², the triangle OT₁P is right-angled at T₁.

Answer: ∠OT₁P = 90°. The tangent PT₁ is perpendicular to the radius OT₁ at the point of tangency. Confirmed.

Example 9: Constructing Tangent to a Circle of Given Radius

Problem: Draw a circle of radius 3.5 cm. From a point P at 6 cm from the centre, construct tangents. Also find ∠T₁OT₂.


Solution:

Given:

  • r = 3.5 cm, d = 6 cm

Steps:

  1. Draw circle, centre O, radius 3.5 cm.
  2. Mark P at 6 cm from O.
  3. Perpendicular bisector of OP gives M. OM = 3 cm.
  4. Auxiliary circle: centre M, radius 3 cm.
  5. Intersection points T₁, T₂ with original circle.
  6. Join PT₁, PT₂.

Tangent length: √(36 − 12.25) = √23.75 ≈ 4.87 cm

∠T₁OT₂:

  • cos(∠T₁OP) = OT₁/OP = 3.5/6 ≈ 0.583
  • ∠T₁OP ≈ 54.3°
  • ∠T₁OT₂ = 2 × 54.3° ≈ 108.6°

Example 10: Special Case: Point on the Circle (d = r)

Problem: Can we draw tangents from a point that lies ON the circle? How many?


Solution:

Analysis:

  • When d = r, the point P lies on the circle.
  • From a point on the circle, exactly one tangent can be drawn.
  • This tangent is perpendicular to the radius at that point.

Construction:

  1. Join OP (which is the radius, since P is on the circle).
  2. At P, construct the perpendicular to OP.
  3. This perpendicular is the tangent.

Note: The auxiliary circle method (for external points) does not apply here because d = r means the auxiliary circle collapses to a point.

Real-World Applications

Engineering — Gear Design:

  • Tangent lines to circular gears determine the contact line between meshing gears.
  • Belt drives wrap around pulleys along tangent lines.

Optics — Light Rays:

  • A light ray tangent to a curved mirror or lens surface determines the angle of reflection/refraction.

Navigation:

  • A ship sailing past a circular island follows a tangent path to avoid the island.
  • The closest approach distance equals the tangent length calculation.

Road Design:

  • A straight road meeting a circular roundabout enters along a tangent to the roundabout.

Astronomy:

  • The line of sight from Earth to the edge of a planet is a tangent to the planet's circular cross-section.

Key Points to Remember

  • A tangent touches a circle at exactly one point and is perpendicular to the radius there.
  • From a point on the circle: draw the radius, then construct the perpendicular. This gives one tangent.
  • From an external point: use the midpoint of OP and an auxiliary circle. This gives two tangents.
  • The two tangents from an external point are equal in length: √(d² − r²).
  • The auxiliary circle works because the angle in a semicircle is 90°, ensuring the tangent ⊥ radius condition.
  • The angle between two tangents from an external point depends on r and d.
  • When d = 2r, the angle between tangents is 60°.
  • As the external point moves farther away, the tangent length increases and the angle between tangents decreases.
  • Always verify by checking ∠OTP = 90° (using Pythagoras).
  • This construction is a high-weightage question in CBSE board exams (4–5 marks).

Practice Problems

  1. Draw a circle of radius 5 cm. Construct a tangent at any point on the circle.
  2. Draw a circle of radius 4 cm. From a point 7 cm from the centre, construct two tangents and measure their lengths.
  3. Draw a circle of radius 3 cm. From a point at distance 5 cm from the centre, construct tangents. Verify that the tangent length is 4 cm.
  4. Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle.
  5. From a point at distance 10 cm from the centre of a circle of radius 6 cm, construct tangents. Calculate the expected tangent length before measuring.
  6. Construct tangents from a point at distance 8 cm from the centre of a circle with radius 4 cm. Find the angle between the two tangents.
  7. Draw a circle of radius 2.5 cm. Mark a point at 6 cm from the centre. Construct the tangents and measure ∠T₁PT₂.
  8. Prove by construction that the two tangents from an external point to a circle are equal in length.

Frequently Asked Questions

Q1. Why does the auxiliary circle intersect the original circle at the points of tangency?

The auxiliary circle has OP as diameter. Any point T on this circle satisfies ∠OTP = 90° (angle in semicircle). If T also lies on the original circle, then OT is a radius and PT ⊥ OT, making PT a tangent by definition.

Q2. How many tangents can be drawn from a point inside the circle?

Zero. A point inside the circle cannot have any tangent to the circle because every line through an interior point intersects the circle at two points (it is a secant, not a tangent).

Q3. Why are the two tangents from an external point equal?

In triangles OT₁P and OT₂P: OT₁ = OT₂ (radii), OP is common, ∠OT₁P = ∠OT₂P = 90°. By RHS congruence, the triangles are congruent. Therefore PT₁ = PT₂.

Q4. What if I cannot find the midpoint of OP accurately?

Use the standard perpendicular bisector construction carefully: with O and P as centres, draw arcs with the same radius (more than half of OP) on both sides. The line through the intersection points bisects OP at M.

Q5. Can this construction be done without the auxiliary circle?

Not easily for tangents from an external point. The auxiliary circle method is the standard NCERT approach. Without it, you would need to calculate angles or distances, which defeats the purpose of a ruler-and-compass construction.

Q6. What is the relationship between ∠T₁PT₂ and ∠T₁OT₂?

They are supplementary: ∠T₁PT₂ + ∠T₁OT₂ = 180°. This is because quadrilateral OT₁PT₂ has two right angles (at T₁ and T₂), and the sum of angles in a quadrilateral is 360°.

Q7. Is the tangent construction asked every year in board exams?

It is one of the most frequently asked construction questions. Typical problems specify a radius and distance from centre, and ask to construct the tangents, measure their lengths, and sometimes find the angle between them.

Q8. How do I draw a tangent if only the circle is given (centre not marked)?

First find the centre by drawing two non-parallel chords and their perpendicular bisectors. The bisectors intersect at the centre. Then proceed with the standard construction.

Q9. What is the angle in a semicircle and why is it used here?

The angle subtended by a diameter at any point on the circle is 90° — this is Thales' theorem. In the construction, the auxiliary circle has OP as diameter, so any point on it (including T₁, T₂) forms a 90° angle with O and P, ensuring the tangent condition.

Q10. Can I construct tangents to an ellipse using this method?

No. This method is specific to circles because it relies on the property that all radii are equal and the angle-in-semicircle theorem. Tangent construction for ellipses requires different techniques.

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