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Tangents from an External Point and Exercises

🎓 Class 10 Mathematics CBSE Theory Ch 10 — Circles ⏱ ~30 min
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This MCQ module is based on: Tangents from an External Point and Exercises

This mathematics assessment will be based on: Tangents from an External Point and Exercises
Targeting Class 10 level in Geometry, with Intermediate difficulty.

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10.3 Number of Tangents from a Point on a Circle

How many tangents can be drawn to a circle from a point? The answer depends on where the point lies — inside, on, or outside the circle.

P (i) P inside: 0 tangents P (ii) P on circle: 1 tangent P T₁ T₂ (iii) P outside: 2 tangents
Fig. 10.6 — Cases (i), (ii) and (iii) for a point's position w.r.t. a circle
  • Case 1. P lies inside the circle ⇒ every line through P is a secant ⇒ no tangent from P.
  • Case 2. P lies on the circle ⇒ exactly one tangent — the one at the point P itself.
  • Case 3. P lies outside the circle ⇒ exactly two tangents from P (one on each side of the line joining P to the centre).
Length of a Tangent
The length of the tangent from an external point P to a circle is the distance between P and the point of contact of that tangent — e.g. if PT is a tangent and T the contact point, the length is PT.

Theorem 10.2 — Tangents from an External Point

Theorem 10.2
The lengths of tangents drawn from an external point to a circle are equal.
O P Q R
Fig. 10.7 — Two tangents PQ and PR from external point P; radii OQ and OR

Proof. Given a circle with centre O, an external point P and two tangents PQ and PR from P with points of contact Q and R. Join OP, OQ and OR. We prove PQ = PR.

In △OQP and △ORP:

  • OQ = OR (radii of the same circle)
  • ∠OQP = ∠ORP = 90° (by Theorem 10.1, tangent ⊥ radius)
  • OP = OP (common side)

Hence △OQP ≅ △ORP by RHS congruence. Therefore PQ = PR (CPCT). \(\blacksquare\)

Consequences
  1. The two tangents subtend equal angles at the centre: ∠QOP = ∠ROP.
  2. The two tangents are equally inclined to the line joining the external point to the centre: ∠OPQ = ∠OPR.
  3. Hence OP is the perpendicular bisector of the chord of contact QR.

Example 1 — Quadrilateral with inscribed circle

Prove that in any parallelogram circumscribing a circle (i.e. all four sides are tangents), the parallelogram is a rhombus.

A B C D P Q R S
Fig. 10.8 — Parallelogram ABCD with inscribed circle, points of contact P, Q, R, S

Solution. Let ABCD be a parallelogram circumscribing a circle, touching AB at P, BC at Q, CD at R, DA at S. By Theorem 10.2, tangent lengths from each vertex are equal:

\[AP=AS,\ BP=BQ,\ CQ=CR,\ DR=DS.\]

Adding: (AP + BP) + (CQ + DQ... wait more carefully) AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = AS + DS + BQ + CQ = AD + BC.

So AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, giving 2AB = 2AD, hence AB = AD. All four sides are equal ⇒ ABCD is a rhombus. \(\blacksquare\)

Example 2 — Tangent length calculation

Two concentric circles have radii 5 cm and 3 cm. Find the length of a chord of the larger circle that touches the smaller circle.

Solution. Let O be the common centre, AB the chord of the larger circle tangent to the smaller one at P. Then OP = 3 cm (radius of inner), OA = 5 cm (radius of outer) and OP ⊥ AB by Theorem 10.1. OP bisects AB.

\[AP=\sqrt{OA^2-OP^2}=\sqrt{25-9}=4\ \text{cm}.\] \[AB=2\times AP=8\ \text{cm}.\]

Example 3 — Triangle tangent lengths

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

A B C D F E 8 6
Fig. 10.9 — △ABC with inscribed circle touching BC at D, CA at E, AB at F

Solution. Let BF = BD = 8, CE = CD = 6, AE = AF = x (equal tangents). Then AB = x + 8, AC = x + 6, BC = 14, and we can let r = 4 cm be the inradius.

Area of △ABC by two methods: Using r and semi-perimeter s = (AB + BC + CA)/2 = (x + 8 + 14 + x + 6)/2 = x + 14: Area = rs = 4(x + 14). Using Heron's formula: Area = √[s(s−a)(s−b)(s−c)] = √[(x+14)(x)(8)(6)] = √[48x(x+14)].

Squaring: 16(x+14)² = 48x(x+14) ⇒ (x+14) = 3x ⇒ x = 7.

So AB = 15 cm and AC = 13 cm.

Activity — Tangent-lengths in a quadrilateral

Draw a quadrilateral ABCD circumscribing a circle. Mark the points of contact on AB, BC, CD, DA. Measure each of the eight tangent-segments. Add AB + CD. Add AD + BC. You will always get the same answer. Try this claim for several random circumscribing quadrilaterals.

By Theorem 10.2, tangent pairs from each vertex are equal. Summing opposite pairs: AB + CD = AD + BC. This is the classical Pitot theorem for tangential quadrilaterals.

Exercise 10.2

Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is: (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm.
r² = 25² − 24² = 625 − 576 = 49 ⇒ r = 7 cm, option (A).
Q2. In the figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ equals: (A) 60° (B) 70° (C) 80° (D) 90°.
Quadrilateral OPTQ has ∠OPT = ∠OQT = 90°. Sum = 360° ⇒ ∠PTQ = 360° − 90° − 90° − 110° = 70°, option (B).
Q3. If tangents PA and PB from a point P to a circle with centre O are inclined at an angle of 80°, then ∠POA is equal to: (A) 50° (B) 60° (C) 70° (D) 80°.
∠APO = 40° (half of 80°). In right △OAP: ∠POA = 90° − 40° = 50°, option (A).
Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let AB be a diameter. Tangent at A ⊥ OA, tangent at B ⊥ OB. Since OA and OB lie along the same straight line AB, both tangents are perpendicular to the same line — hence parallel to each other.
Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Let tangent at P be ℓ. The perpendicular to ℓ at P is unique. By Theorem 10.1, OP is perpendicular to ℓ at P. Therefore OP is exactly this perpendicular; it passes through the centre.
Q6. The length of a tangent from a point A at distance 5 cm from the centre of a circle is 4 cm. Find the radius of the circle.
r² = 5² − 4² = 9 ⇒ r = 3 cm.
Q7. Two concentric circles of radii 5 cm and 3 cm. Find length of the chord of the larger circle that touches the smaller circle.
Half-chord = √(25 − 9) = 4. Full chord = 8 cm.
Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Let P, Q, R, S be points of contact on AB, BC, CD, DA. By equal-tangent theorem: AP = AS, BP = BQ, CQ = CR, DR = DS. Adding: AP + BP + CR + DR = AS + BQ + CQ + DS ⇒ AB + CD = AD + BC.
Q9. In the given figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.
Let tangents XY, X'Y' touch the circle at P and Q. OA bisects ∠PAC, OB bisects ∠QBC (by equal-tangent symmetry). Also ∠PAC + ∠QBC = 180° (co-interior angles of parallel lines XY and X'Y' cut by transversal AB). Hence ∠OAC + ∠OBC = 90°. In △AOB, ∠AOB = 180° − 90° = 90°.
Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
In quadrilateral OPTQ (where PT, QT are tangents with O the centre), ∠OPT = ∠OQT = 90°. Sum of angles = 360° ⇒ ∠POQ + ∠PTQ = 180°, i.e. supplementary.
Q11. Prove that the parallelogram circumscribing a circle is a rhombus.
From Example 1: AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, so 2AB = 2AD ⇒ AB = AD. All four sides equal ⇒ rhombus.
Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that BD = 8 cm, DC = 6 cm (D on BC). Find AB and AC.
As worked in Example 3: x = 7, so AB = 15 cm, AC = 13 cm.
Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.
Let quadrilateral ABCD circumscribe a circle, points of contact P, Q, R, S on AB, BC, CD, DA. Join OA, OB, OC, OD. From congruent right-triangle pairs, ∠AOS = ∠AOP, ∠BOP = ∠BOQ, ∠COQ = ∠COR, ∠DOR = ∠DOS. Total around O = 360°. Grouping: (∠AOS+∠AOP)+(∠BOP+∠BOQ)+(∠COQ+∠COR)+(∠DOR+∠DOS) = 360° ⇒ 2(∠AOP+∠BOQ+∠COR+∠DOS) = 360° ⇒ ∠AOP+∠BOQ+∠COR+∠DOS = 180°. Angle subtended by AB at centre = ∠AOB = ∠AOP + ∠POB, and similarly for CD. Adding opposite pairs ⇒ 180°, supplementary.

10.4 Summary

Key Points
  1. A tangent to a circle is a line that meets the circle at exactly one point — the point of contact.
  2. Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  3. Theorem 10.2: The lengths of the two tangents drawn from an external point to a circle are equal.
  4. From an external point there are exactly two tangents; from a point on the circle, exactly one; from an interior point, zero.
Looking Ahead
In the next chapter — Areas Related to Circles — you will use these ideas together with the earlier formula for circumference and area to calculate areas of sectors, segments and complex combinations of plane figures.
Competency-Based Questions
Q1. A bicycle-repair shop must cut a leather patch in the shape of a rhombus to circumscribe a circular coin of radius 3 cm. Apply Theorem 10.2 to determine what side-length makes the rhombus smallest while still circumscribing the coin.
L3 Apply
The smallest such rhombus is actually the square (rhombus with 90° angles); its side equals the diameter = 6 cm. Any non-square rhombus circumscribing the same circle has a longer side, because it must still have in-radius 3.
Q2. Analyse why the two tangents from an external point are symmetric about the line joining the point to the centre.
L4 Analyse
Both right triangles △OQP and △ORP are congruent (RHS: OQ = OR radii, OP common, ∠OQP = ∠ORP = 90°). Corresponding angles at P are equal ⇒ line OP bisects ∠QPR. Symmetry follows from reflecting across OP — tangent QP maps to RP and vice versa.
Q3. Evaluate the statement: "A circle inscribed in an equilateral triangle always has centre coinciding with the triangle's centroid."
L5 Evaluate
True. In an equilateral triangle, the incentre (centre of inscribed circle) and centroid coincide — also with circumcentre and orthocentre. This follows from the triangle's full symmetry: all medians, angle bisectors and perpendicular bisectors meet at the same point.
Q4. Create a real-world problem involving two wheels with a tangent belt connecting them, and compute the length of the belt for wheels of radii 8 cm and 3 cm whose centres are 13 cm apart (cross belt configuration).
L6 Create
Problem: "Two pulleys of radii 8 cm and 3 cm have centres 13 cm apart. The pulleys are connected by a crossed belt that wraps around each. Compute the length of the straight portion of each tangent section." Solution: For a cross-belt, tangent length between circles = √(d² − (r₁ + r₂)²) = √(169 − 121) = √48 = 4√3 ≈ 6.93 cm.
Assertion & Reason
A: Tangents drawn at the two endpoints of a diameter of a circle are parallel.
R: Both tangents are perpendicular to the same diameter.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. Two lines perpendicular to the same line are parallel.
A: A quadrilateral circumscribing a circle always satisfies AB + CD = AD + BC.
R: This is the Pitot theorem, proved using the equality of tangent lengths from each external vertex.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. Standard application of Theorem 10.2.
A: If two tangents from a point P to a circle are perpendicular, the distance OP equals r√2.
R: OPQR (O centre, Q, R contact points) is a square of side r when ∠QPR = 90°.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. OQPR has four right angles and OQ = OR = r; also PQ = PR by Theorem 10.2. With ∠QPR = 90°, it's a square of side r, diagonal OP = r√2.

Frequently Asked Questions

Why are tangents from an external point equal?
Triangles formed by the two tangents, the two radii and the line joining the external point to the centre are congruent by RHS, giving equal tangent lengths.
What is the summary of Class 10 Chapter 10 Circles?
Chapter 10 covers tangents to circles: their definition, the perpendicular-to-radius property (Theorem 10.1), and the equal-length property from an external point (Theorem 10.2).
How do you find the length of a tangent from an external point?
Use the right triangle formed by the radius, tangent and line to centre: tangent length = root( OP^2 - r^2 ), where OP is the distance to the centre and r the radius.
What is a common mistake in circle tangent problems?
Forgetting to mark the right angle at the point of contact, or not using the RHS congruence when proving equal tangents.
Are tangent problems common in CBSE boards?
Yes. Tangent properties are a regular 3-4 mark proof/construction question in Class 10 CBSE board exams and appear in most previous years' papers.
How are tangents used in constructions?
Chapter 11 of Class 10 uses tangent properties to construct tangents from an external point using the midpoint-and-semicircle method.
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