Exercise 9.2 — Construction of Tangents

Construction of pair of tangents from an external point.

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Exercise 9.2 — Tangents and Secants to a Circle

Exercise 9.2 from the chapter Tangents and Secants to a Circle is one of the most scoring chapters in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). It builds on the basic idea of a tangent — a line that touches a circle at exactly one point — and uses it to solve numerical problems, geometric proofs, and ruler-and-compass constructions that frequently appear in board exams.

This exercise has nine questions in total: one multiple-choice question with five parts, three numerical/proof-based problems, and five construction problems. Every single one of them rests on just two core ideas — the number of tangents possible from a point, and the fact that the two tangents drawn from any external point to a circle are always equal in length. Master these two ideas and the entire exercise becomes straightforward.

Number of Tangents Equal Tangent Lengths MCQ with Reasoning Geometric Proofs Ruler-Compass Constructions

💡 Foundation fact: A tangent to a circle is always perpendicular to the radius drawn at the point where it touches the circle. Almost every proof in this exercise begins by marking a 90° angle at the point of contact — keep this in mind as you work through each question.

How Many Tangents Can Be Drawn From a Point?

Before solving any problem in this exercise, it is essential to know how many tangents can be drawn to a circle depending on where the point is located — inside the circle, on the circle, or outside it. This single idea is tested directly in Question 9 and used indirectly in almost every other question.

Point inside the circle

0 tangents possible — every line through P cuts the circle at two points

Point on the circle

Exactly 1 tangent possible — perpendicular to radius OP

Point outside the circle

Exactly 2 tangents possible, equal in length

Inside the circle: Zero tangents can be drawn — any straight line through such a point intersects the circle at two points, making it a secant, never a tangent.
On the circle: Exactly one tangent can be drawn, and it is always perpendicular to the radius at that point.
Outside the circle: Exactly two tangents can be drawn, and — as proved in the next section — both tangents are always equal in length.

Theorem — Tangents From an External Point Are Equal

This is the single most important result in the whole exercise. It states that if two tangents are drawn from the same external point to a circle, their lengths are always equal. Six out of the nine questions in Exercise 9.2 use this theorem directly.

PA and PB are tangents from external point P — by the theorem, PA = PB

Given: A circle with centre O, and P is a point lying outside the circle. PA and PB are the two tangents drawn from P, touching the circle at A and B.
To prove: PA = PB

Join OA, OB and OP. ∠OAP = ∠OBP = 90° (tangent ⟂ radius at point of contact) In right triangles ΔOAP and ΔOBP: OA = OB (radii of the same circle) OP = OP (common side) ⟹ ΔOAP ≅ ΔOBP (by RHS congruence rule) ⟹ PA = PB (by CPCT)

Tangent length from external point P = √(OP² − r²), where r is the radius and OP is the distance from the centre

📌 Why this matters: This formula — derived directly from the Pythagoras theorem in the right triangle OAP — is the engine behind almost every numerical problem in this exercise, including Questions 1(ii), 2, 4, 5 and 6.

Question 1 — Multiple Choice with Justification

This question has five sub-parts, each testing a different application of the two core ideas above. In board exams, marks are awarded not just for choosing the right option, but for the reasoning that justifies it — so the "why" matters as much as the "what."

#	Question (short)	Correct Option	Reasoning
(i)	Angle between a tangent and the radius at the point of contact	90°	By definition, a tangent at any point of a circle is always perpendicular to the radius drawn to that point.
(ii)	Tangent length = 24 cm, distance of point from centre = 25 cm; find the radius	7 cm	Using the right triangle formed by the radius, tangent and the line to the centre: 25² = r² + 24², so r² = 625 − 576 = 49, giving r = 7 cm.
(iii)	AP, AQ are tangents with ∠POQ = 110°; find ∠PAQ	70°	In quadrilateral OPAQ, the angles at P and Q are 90° each (tangent ⟂ radius). Since all four angles sum to 360°, ∠PAQ = 360° − 90° − 90° − 110° = 70°.
(iv)	Tangents PA, PB inclined at 80° to each other; find ∠POA	50°	OP bisects ∠APB (since ΔOAP ≅ ΔOBP), so ∠APO = 40°. In ΔOAP, ∠OAP = 90°, so ∠POA = 180° − 90° − 40° = 50°.
(v)	Two parallel tangents with a third tangent AB joining them at C; find ∠AOB	90°	Using equal tangent lengths and right angles at the points of contact, OPAC forms a square, giving ∠COA = 45° and, by symmetry, ∠COB = 45°, so ∠AOB = 90°.

⚠️ Common exam trap: In part (iii) and (v), many students forget that the quadrilateral formed by the centre, the two tangent points, and the external point has two right angles already built in — always mark these 90° angles first before trying to find the unknown angle.

Question 2 — Chord of a Larger Circle Touching a Smaller Concentric Circle

Two concentric circles (circles sharing the same centre) of radii 5 cm and 3 cm are drawn. We need to find the length of the chord of the larger circle that just touches — is tangent to — the smaller circle.

AB is a chord of the larger circle, tangent to the smaller circle at P

Solution

Find the length of chord AB

OP ⟂ AB (radius at the point of tangency), OP = 3 cm, OA = 5 cm In right triangle OPA: OA² = OP² + AP² 5² = 3² + AP² ⟹ AP² = 25 − 9 = 16 AP = 4 cm AB = 2 × AP (perpendicular from the centre bisects the chord) AB = 2 × 4 = 8 cm

✅ Answer: The length of the chord of the larger circle that touches the smaller circle is 8 cm.

Question 3 — A Parallelogram Circumscribing a Circle is a Rhombus

This question asks us to prove that if all four sides of a parallelogram touch a circle (the circle is inscribed inside it), the parallelogram must actually be a rhombus — a parallelogram with all sides equal.

Parallelogram ABCD touches the circle at P, Q, R and S

Proof

Show that AB = AD, so ABCD is a rhombus

AB, BC, CD and AD touch the circle at P, Q, R and S respectively. By the equal-tangent-length theorem: AP = AS BP = BQ CR = CQ DR = DS Adding all four: (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) ⟹ AB + CD = AD + BC Since ABCD is a parallelogram, AB = CD and AD = BC, so: 2·AB = 2·AD ⟹ AB = AD

✅ Conclusion: Since adjacent sides AB and AD of the parallelogram are equal, all four sides must be equal — therefore the parallelogram is a rhombus.

Question 4 — Finding the Sides of a Triangle Circumscribing a Circle

Triangle ABC circumscribes a circle of radius 3 cm. The point of contact D on side BC divides it into BD = 9 cm and DC = 3 cm. We need to find the lengths of the other two sides, AB and AC.

Incircle touches BC at D, AB at E and AC at F (right angle at C, since AC ⟂ BC)

Solution

Find AB and AC

CD = CF = 3 cm, BD = BE = 9 cm (equal tangents from C and B) Let AE = AF = x. Then AB = x + 9, AC = x + 3, BC = 12 OECD is a square (OE = OD = radius, ∠E = ∠D = 90°), so ∠C = 90° Triangle ACB is right-angled at C, so by Pythagoras: AB² = AC² + BC² (x+9)² = (x+3)² + 12² x² + 18x + 81 = x² + 6x + 9 + 144 12x = 72 ⟹ x = 6 AB = 6 + 9 = 15 cm AC = 6 + 3 = 9 cm

✅ Answer: AB = 15 cm and AC = 9 cm.

📌 Pattern to remember: Whenever a circle is inscribed in a triangle, label the two equal tangent segments from each vertex with the same variable (here, x for vertex A) — this turns the problem into a simple algebraic equation.

Construction Questions — Drawing Tangent Pairs (Q5, Q6, Q7 & Q9)

Four questions in this exercise (5, 6, 7 and 9) ask you to actually construct tangents using a ruler and compass, rather than just calculate. All of them use exactly the same construction method — only the circle's radius and the distance of the external point change.

The standard tangent-pair construction: L is the midpoint of OP

The General Construction Method

Draw the given circle with centre O and the given radius.
Mark the external point P at the given distance from O, and join OP.
Draw the perpendicular bisector of OP — let it cut OP at the midpoint L.
With L as centre and LO (or LP) as radius, draw a second, larger circle.
This new circle intersects the original circle at two points, A and B.
Join PA and PB — these are the required tangents.

💡 Why this works: Since OP is a diameter of the larger auxiliary circle, the angle ∠OAP (and ∠OBP) is automatically 90° — the angle in a semicircle is always a right angle. A line meeting the radius OA at exactly 90° at point A must be a tangent, so PA is guaranteed to be a tangent without any guesswork.

Question	Setup	Tangent Length Found	Verification
Q5	Circle of radius 6 cm; external point 10 cm from centre	8 cm	6² + 8² = 36 + 64 = 100 = 10² ✓ Pythagoras theorem verified
Q6	Circle of radius 4 cm; point taken on a concentric circle of radius 6 cm	≈ 4.5 cm	6² − 4² = 36 − 16 = 20 ≈ 4.5² ✓ Pythagoras theorem verified
Q7	Circle drawn using a bangle (centre found using two chords and their perpendicular bisectors); external point chosen freely	PG = PH	By measurement, both tangents from the chosen point are exactly equal in length
Q9	Circle with centre O; external point R	2 tangents	Confirms the theory from Section 2: any external point allows exactly two tangents

Notice the clever trick in Question 7: since no centre is marked on a circle traced from a bangle, you must first locate the centre by drawing two chords and constructing their perpendicular bisectors — the bisectors always meet exactly at the centre of the circle. Once the centre is found, the same six-step method above applies.

Question 8 — Tangent at P Bisects Side BC

In a right triangle ABC, a circle is drawn with side AB as its diameter, intersecting the hypotenuse AC at point P. We must prove that the tangent drawn to the circle at P bisects the third side, BC.

The tangent through P meets BC at Q, the midpoint of BC

Proof

Show that Q is the midpoint of BC

In ΔQPC, ∠QPC = ∠QCP (given/angle in alternate segment) ⟹ QP = QC (sides opposite equal angles) ...(1) Also, QP = QB (tangents from external point Q are equal) ...(2) From (1) and (2): QC = QB

✅ Conclusion: Since QC = QB, the point Q — where the tangent at P meets BC — is exactly the midpoint of BC. Hence proved.

📌 Key insight: This proof elegantly combines two ideas from earlier in the chapter: the tangent-chord angle property (∠QPC = ∠QCP) and the equal-tangent-length theorem (QP = QB) — showing how every result in this exercise connects back to the same handful of core facts.

Common Mistakes to Avoid

Forgetting the 90° angle: Many students lose marks by not explicitly marking ∠OAP = 90° (tangent ⟂ radius) before starting Pythagoras-based calculations.
Mixing up OP² = OA² + AP² with OA² = OP² + AP²: Always identify which side is the hypotenuse — it is the line joining the centre to the external point, since the right angle is at the point of contact, not at the centre.
Assuming a square without proving it: In Question 1(v) and Question 4, the fact that a quadrilateral is a square (e.g. OECD) must be justified by stating equal radii, equal tangent lengths, and 90° angles — not simply assumed from the figure.
Skipping the RHS congruency reasoning: When proving PA = PB, write out the full RHS (Right angle – Hypotenuse – Side) congruence statement; jumping straight to the conclusion costs marks in board exams.
Construction without the perpendicular bisector step: A frequent slip in Questions 5, 6, 7 and 9 is trying to draw tangents "by eye." Always construct the midpoint L of OP using a proper perpendicular bisector — this is what makes the construction mathematically valid.

⛔ High-value exam tip: Question 1 alone (with its five sub-parts) is frequently repeated in CBSE and Telangana/AP board papers in a slightly modified numeric form. Practising the reasoning behind each part — not just memorising the final angle — prepares you for any variation the exam sets.

Quick Reference — All Answers at a Glance

Question	Topic	Answer / Result
Q1(i)	Angle between tangent and radius	90°
Q1(ii)	Radius from tangent length 24 cm, OQ = 25 cm	7 cm
Q1(iii)	∠PAQ when ∠POQ = 110°	70°
Q1(iv)	∠POA when tangents inclined at 80°	50°
Q1(v)	∠AOB with two parallel tangents and a third tangent	90°
Q2	Chord of larger circle touching smaller concentric circle (r = 5, 3 cm)	8 cm
Q3	Parallelogram circumscribing a circle	Proved to be a rhombus
Q4	Sides AB, AC of triangle with incircle (r=3, BD=9, DC=3)	AB = 15 cm, AC = 9 cm
Q5	Tangent length, r = 6 cm, OP = 10 cm	8 cm
Q6	Tangent length, concentric circles r = 4 cm & 6 cm	≈ 4.5 cm
Q7	Bangle construction, tangent pair from external point	PG = PH (equal tangents)
Q8	Tangent at P on circle (diameter AB) bisects BC	Proved (Q is midpoint of BC)
Q9	Tangents from external point R	2 tangents

What This Lesson Prepares You For

Exercise 9.2 builds directly on the basic definitions covered in the first part of this chapter, where the idea of a tangent and a secant is introduced. If any of the foundational ideas here felt unfamiliar, it helps to revisit the introductory exercise on tangents to a circle before attempting the proofs in this exercise.

The RHS congruence rule used repeatedly in this exercise (to prove PA = PB and similar results) is a core technique from the Triangles chapter, so strengthening your congruence proofs there will make every proof here faster to write. Similarly, the perpendicular-bisector construction technique used in Questions 5, 6, 7 and 9 is the same skill first introduced while learning the construction of quadrilaterals in earlier classes.

📐 Board Exam Tip (CBSE, Telangana & AP): The theorem "tangents drawn from an external point are equal" is one of the most frequently asked 2-mark to 4-mark proof questions across all three boards. Practise writing the full proof — Given, To Prove, Construction, and Proof steps — at least three or four times until it becomes automatic.