Exercise 12.3 — Perpendiculars from Centre
Perpendiculars from centre to a chord and three points describing a circle.
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Key Theorems for Exercise 12.3
All five problems in this exercise are built on four foundational theorems about circles and chords. Understanding these theorems — rather than just memorising them — is the key to solving construction and proof questions with confidence.
Theorem 1 — Perpendicular from Centre
The perpendicular from the centre of a circle to a chord bisects the chord. If OM ⊥ AB then AM = BM.
Theorem 2 — Converse
The line drawn through the centre of a circle to bisect a chord is perpendicular to that chord. If AM = BM then OM ⊥ AB.
Theorem 3 — Circumcircle
There is one and only one circle passing through three non-collinear points. The perpendicular bisectors of any two sides of a triangle meet at the circumcentre.
Theorem 4 — Equal Chords
Equal chords of a circle are equidistant from the centre, and conversely, chords at equal distances from the centre are equal in length.
OM ⊥ AB → AM = BM
AB = CD ⟺ OM = ON
🔑 Circumcircle key fact: The circumcentre (where perpendicular bisectors meet) is equidistant from all three vertices of the triangle. This equal distance is the circumradius. The circumcentre can lie inside, outside, or on the triangle depending on whether it is acute, obtuse, or right-angled.
Question 1 — Construct Circumcircles of Triangles
The circumcircle of a triangle is the unique circle that passes through all three vertices. To construct it, we use the fact that the circumcentre lies at the intersection of the perpendicular bisectors of any two sides.
Circumcentre S = Intersection of perpendicular bisectors of any two sides
Circumradius = SA = SB = SC (distance from S to any vertex)
Question 1 (i)
In △ABC, AB = 6 cm, BC = 7 cm and ∠A = 60°. Construct the circumcircle.
| Given | AB = 6 cm, BC = 7 cm, ∠A = 60° |
| To Construct | Circumcircle of △ABC |
| Key Idea | Perpendicular bisectors of two sides meet at circumcentre S |
Circumcircle of △ABC (AB=6, BC=7, ∠A=60°)
Steps of Construction
- Step 1: Construct △ABC with the given measurements — draw AB = 6 cm, construct ∠A = 60°, and mark C such that BC = 7 cm.
- Step 2: Draw the perpendicular bisector of side AB using a compass and ruler (open compass to more than half of AB, draw arcs from A and B, join the two intersection points).
- Step 3: Draw the perpendicular bisector of side BC in the same way.
- Step 4: Mark the intersection point of these two perpendicular bisectors as 'S'. This is the circumcentre.
- Step 5: Taking S as centre and SA as radius (= SB = SC), draw a circle.
- Step 6: This circle passes through all three vertices A, B, and C. This is the circumcircle of △ABC.
Question 1 (ii)
In △PQR, PQ = 5 cm, QR = 6 cm and RP = 8.2 cm. Construct the circumcircle.
| Given | PQ = 5 cm, QR = 6 cm, RP = 8.2 cm (SSS — all three sides given) |
| To Construct | Circumcircle of △PQR |
Circumcircle of △PQR (SSS triangle)
- Step 1: Construct △PQR with PQ = 5 cm, QR = 6 cm and RP = 8.2 cm using compass arcs (SSS construction).
- Step 2: Draw the perpendicular bisector of side PQ.
- Step 3: Draw the perpendicular bisector of side QR.
- Step 4: Mark the intersection point as 'S' (circumcentre).
- Step 5: With S as centre and SP as radius, draw a circle passing through Q and R.
- Step 6: This is the circumcircle of △PQR.
Question 1 (iii)
In △XYZ, XY = 4.8 cm, ∠X = 60° and ∠Y = 70°. Construct the circumcircle.
| Given | XY = 4.8 cm, ∠X = 60°, ∠Y = 70° (ASA — two angles and included side) |
| Note | ∠Z = 180° − 60° − 70° = 50° |
| To Construct | Circumcircle of △XYZ |
Circumcircle of △XYZ (ASA triangle)
- Step 1: Construct △XYZ — draw XY = 4.8 cm, construct ∠X = 60° and ∠Y = 70°. The arms intersect at Z.
- Step 2: Draw the perpendicular bisector of side XY.
- Step 3: Draw the perpendicular bisector of side YZ.
- Step 4: Mark the intersection point as 'S'.
- Step 5: With S as centre and SX as radius, draw a circle through Y and Z.
- Step 6: This is the circumcircle of △XYZ.
💡 All three sub-parts follow identical steps — only the given measurements differ. The universal circumcircle method always involves: construct the triangle → draw two perpendicular bisectors → their intersection is S (circumcentre) → draw circle with radius = S to any vertex.
Question 2 — Two Circles Through Two Points A and B
Question 2
Draw two circles passing through A and B where AB = 5.4 cm.
| Given | Line segment AB = 5.4 cm |
| To Draw | Two different circles, each passing through both A and B |
| Key Principle | Infinitely many circles can pass through two given points; centres must lie on the perpendicular bisector of AB |
Two circles — both pass through A and B
- Step 1: Draw a line segment AB = 5.4 cm using a ruler.
- Step 2: Set compass to 5.4 cm. Place at A and draw a circle — this circle passes through B. (Centre = A, radius = AB)
- Step 3: With the same radius (5.4 cm) and B as centre, draw another circle — this also passes through A. (Centre = B, radius = BA)
✅ The two circles overlap, forming a "lens" shape called a vesica piscis. Both circles pass through both A and B since the radius equals the distance AB. Any other circle through both A and B would have its centre on the perpendicular bisector of AB.
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Question 3 — Centres Lie on Perpendicular Bisector of Common Chord
Question 3
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
| Given | Two circles with centres M and N intersecting at A and B. AB is the common chord. |
| To Prove | MN is the perpendicular bisector of AB (i.e., MN ⊥ AB and O bisects AB) |
| Construction | Join MA, MB, NA, NB. Let MN and AB intersect at O. |
MN ⊥ AB and OA = OB (perpendicular bisector)
Proof — Part 1: Establish ∠AMN = ∠BMN
Consider triangles △MAN and △MBN:
1MA = MB (radii of circle M)
2NA = NB (radii of circle N)
3MN = MN (common side)
∴By SSS: △MAN ≅ △MBN → ∠AMN = ∠BMN (CPCT)
So ∠AMO = ∠BMO …(1)
Proof — Part 2: Establish OA = OB and ∠AOM = 90°
Now consider triangles △MAO and △MBO:
1MA = MB (radii of circle M)
2MO = MO (common side)
3∠AMO = ∠BMO (from step 1 above)
∴By SAS: △MAO ≅ △MBO
By CPCT: OA = OB …(2) (O bisects AB)
Also by CPCT: ∠AOM = ∠BOM
But ∠AOM + ∠BOM = 180° (linear pair on straight line AB)
∴ 2∠AOM = 180° → ∠AOM = 90° …(3)
Conclusion
From (2): O bisects AB → OA = OB
From (3): MN ⊥ AB → ∠AOM = 90°
∴ MN is the perpendicular bisector of AB.
∴ Centres of two intersecting circles lie on the perpendicular bisector of their common chord. ✓
✓ Proved
📌 Two-step strategy: First prove the two triangles formed by the two centres and the two intersection points are congruent (SSS). Extract the equal angles. Then prove the smaller triangles congruent (SAS) to get OA = OB and ∠AOM = 90°.
Question 4 — Equal Angles with Diameter ⟹ Equal Chords
Question 4
If two intersecting chords of a circle make equal angles with the diameter through their point of intersection, prove that the chords are equal.
| Given | Circle with centre O. Chords AB and CD intersect at E on diameter PQ. OL ⊥ AB and OM ⊥ CD. ∠OEL = ∠OEM. |
| To Prove | AB = CD |
| Key Theorem | Equal chords are equidistant from the centre (and converse). |
∠OEL = ∠OEM → OL = OM → AB = CD
Proof
Consider △OEL and △OEM:
1∠OLE = ∠OME = 90° (OL ⊥ AB, OM ⊥ CD by construction)
2∠OEL = ∠OEM (given — equal angles with diameter)
3OE = OE (common hypotenuse)
∴By AAS: △OEL ≅ △OEM
By CPCT: OL = OM
i.e., distances from centre O to chords AB and CD are equal
By Theorem 4 (chords at equal distance from centre are equal):
∴ AB = CD ✓
✓ AB = CD Proved
Question 5 — Diameter Perpendicular to Chord: AD = BD
Question 5
AB is a chord of a circle with centre O. CD is the diameter perpendicular to AB. Show that AD = BD.
| Given | Circle with centre O. Chord AB. Diameter CD ⊥ AB, intersecting AB at P. |
| To Prove | AD = BD |
| Key Fact | Since CD ⊥ AB, AP = BP (perpendicular from centre bisects the chord) |
CD ⊥ AB → AP = PB → AD = BD
Proof
Since CD is a diameter and CD ⊥ AB, by Theorem 1: the perpendicular from the centre to chord AB bisects it, so AP = BP.
Now consider △APD and △BPD:
1∠APD = ∠BPD = 90° (CD ⊥ AB given)
2AP = BP (perp from centre bisects the chord)
3DP = DP (common side)
∴By SAS: △APD ≅ △BPD
By CPCT: AD = BD ✓
✓ AD = BD Proved
📌 Why SAS works here: The right angle is at P (between the equal sides AP and DP). Both sides adjacent to the right angle are equal in the two triangles — that's the "S-A-S" combination. This is a cleaner path than using RHS, though both would work.
Quick Summary — All Questions at a Glance
| Q | What to Do | Key Tool / Theorem | Result |
|---|---|---|---|
| 1(i) | Circumcircle of △ABC (AB=6, BC=7, ∠A=60°) | Perp bisector of AB & BC; centre S | Circle through A, B, C |
| 1(ii) | Circumcircle of △PQR (SSS: 5, 6, 8.2 cm) | Perp bisector of PQ & QR; centre S | Circle through P, Q, R |
| 1(iii) | Circumcircle of △XYZ (XY=4.8, ∠X=60°, ∠Y=70°) | Perp bisector of XY & YZ; centre S | Circle through X, Y, Z |
| 2 | Draw 2 circles through A & B (AB=5.4 cm) | Circle with centre A, then B; radius = AB | Two overlapping circles |
| 3 | Prove centres M, N lie on perp bisector of chord AB | SSS → SAS → CPCT | OA=OB and MN⊥AB proved |
| 4 | Equal angles with diameter → equal chords | AAS congruence → OL=OM → AB=CD | AB = CD proved |
| 5 | Diameter ⊥ chord AB; show AD = BD | Perp bisects chord → AP=BP, then SAS | AD = BD proved |
Common Mistakes to Avoid in Board Exams
- Drawing only one perpendicular bisector for the circumcircle: You need perpendicular bisectors of at least two sides. One is not enough to fix the circumcentre uniquely.
- Using a vertex (not the circumcentre) as the compass point: After locating S, always measure the radius as the distance from S to a vertex — not the length of a side of the triangle.
- Claiming MN ⊥ AB in Q3 without proving it: You must show both OA = OB (bisects) and ∠AOM = 90° (perpendicular). Missing either part means the perpendicular bisector claim is incomplete.
- Confusing AAS and ASA in Q4: AAS works because we have two angles (90° and the given angle) and a non-included side (OE = OE). Check which two angles and which side match before naming the rule.
- Forgetting to cite Theorem 4 at the end of Q4: After proving OL = OM, you must explicitly state: "Chords equidistant from the centre are equal." Without this, AB = CD is an unsupported jump.
⛔ Most common board exam error in construction questions: Drawing the perpendicular bisector approximately (by eye) instead of using proper compass arcs. Open the compass to more than half the side length, draw arcs from both endpoints, and join the two arc-intersection points — that's the perpendicular bisector. Examiners check for arc marks.
What This Exercise Prepares You For
The circumcircle construction in Q1 directly extends into Exercise 12.4, where you explore the angle subtended by an arc and the relationship between inscribed and central angles. The perpendicular bisector result proved in Q3 is a cornerstone result reused in Class 10 and in coordinate geometry when finding the equation of a circle.
Question 5's result (diameter perpendicular to chord bisects it, making the two chord segments equal, which in turn makes the lines from the far end of the diameter equal) is a special case of the general chord-angle theorem that appears throughout Exercise 12.5 on cyclic quadrilaterals.
For Telangana and Andhra Pradesh board exams, construction questions (Q1 type) typically carry 4 marks, while proof questions (Q3, Q4, Q5 type) carry 4–5 marks each. A clean diagram with all arc marks, labelled points, and a structured "Given → Construction → Proof → Conclusion" layout earns full marks.
✅ Board Exam Strategy (CBSE, Telangana & AP): For proofs, always write the triangles being compared first, then list the three conditions (matching the congruence rule), state the rule, then use CPCT for each required result. Never skip from the congruence rule directly to the final answer — examiners award 1 mark for each step of the structured proof.