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State Board · Class 9 · Mathematics Part 2

Lesson 6: Circle

4 concepts · 32 practice questions with answers · Free to read

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Parts of a circle and chords

From daily life

A bicycle wheel has spokes from the centre to the rim (radii). A stick joining any two points on the rim is a chord. The stick through the centre is the longest, the diameter.

Simple definition

A segment joining any two points of the circle is a chord. A chord through the centre is a diameter; diameter = 2 × radius and it is the longest chord. Joining the centre to a point on the circle gives a radius.

Centre, radius, diameter, chord and central angle of a circleParts of a circleOABCDDiameter AB (longest chord)Chord CDRadius (centre to circle)Central angle ∠COD

In the figure: The diameter AB passes through the centre. The chord CD does not; joining its ends to the centre gives the central angle ∠COD.

Rule 1: Diameter = 2 × radius — Given the radius the diameter is double, and given the diameter the radius is half. Every chord is shorter than or equal to the diameter.

  • Easy: Radius 7 ⇒ diameter 14.
  • Medium: Diameter 13 ⇒ radius 6.5.
  • Exam type: In a circle of radius 6 a chord of 10 and a chord of 12 are both possible (≤ 12). The chord 12 is a diameter. A chord of 13 is impossible.
Live figure

Parts of a circle and chords

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Rule 2: The perpendicular from the centre bisects the chord — If OM ⊥ AB then AM = MB. So the foot of the perpendicular is the midpoint of the chord.

  • Easy: AB = 10 and OM ⊥ AB ⇒ AM = 5.
  • Medium: AM = 7 ⇒ AB = 14.
  • Exam type: Chord AB = 16 and radius 10. AM = 8; OM = √(100 − 64) = 6.

Rule 3: The line from the centre to the midpoint of a chord is perpendicular to it — If AM = MB then OM ⊥ AB, that is ∠OMA = 90°. The reverse is also true: one statement gives the other.

  • Easy: M is the midpoint of chord AB ⇒ ∠OMA = 90°.
  • Medium: In ∆OMA, ∠M = 90° and OA = radius (hypotenuse). OM and AM are the sides forming the right angle.
  • Exam type: OA = OB (radii) and AM = MB ⇒ ∆OAM ≅ ∆OBM (SSS) ⇒ ∠OMA = ∠OMB and together 180° ⇒ each 90°.
Remember

The diameter is the longest chord. A perpendicular from the centre to a chord bisects it.

Common mistake

Mixing up the perpendicular distance and the chord length. OM is from the centre to the chord; chord AB = 2 × AM.

Shortcut

Whenever a chord or distance is given, form ∆OMA with the radius as the hypotenuse. Memory clue: The radius, half the chord and the distance form a right triangle.

Spot the difference

ChordDiameter
Passes through the centre?usually notyes
Lengthless than the diameterlongest, 2 × radius
Perpendicular from the centrebisects the chordthe centre itself is its midpoint

Practice questions

Easy
  1. 1If the radius is 9, what is the diameter?
  2. 2If OM ⊥ AB and AB = 12, what is AM?
  3. 3Which is the longest chord of a circle?
Medium
  1. 4Is a chord of 18 possible in a circle of radius 8?
  2. 5M is the midpoint of chord PQ and O is the centre. Find ∠OMP.
  3. 6If AM = 9, what is the chord AB?
Challenging
  1. 7A chord is drawn at distance 5 from the centre of a circle of radius 13. Find its length.
  2. 8One chord passes through the centre and another is at distance 3 from it; radius 5. Find both lengths.
Show answers (try them yourself first)
  1. 1Diameter = 2 × 9 = 18. 2) AM = 12 ÷ 2 = 6. 3) The diameter.
  2. 4No. The diameter is 16 and no chord can exceed 16. 5) 90°. 6) AB = 2 × 9 = 18.
  3. 7AM = √(169 − 25) = 12, so the chord is 24. 8) The one through the centre is the diameter = 10. The other: AM = √(25 − 9) = 4, chord 8.

Length of a chord and distance from the centre: Pythagoras

From daily life

If a ruler is placed on a round table so that both its ends touch the edge, the farther it is from the centre the shorter it is. It is longest when it passes through the centre.

Simple definition

∆OMB is a right triangle: hypotenuse OB = radius r, sides OM = d (distance) and MB = half the chord. So r² = d² + (half chord)². From any two of the three values we can find the third.

Relation between the perpendicular and the midpoint of a chordThe perpendicular from the centre bisects the chordOABM1351212OM ⊥ AB⇒ AM = MBAM = MB⇒ OM ⊥ ABRight triangle OMB:5² + 12² = 13²chord AB = 24, radius 13

In the figure: Radius 13 and chord 24: half chord 12 and distance 5 (5² + 12² = 13²).

Rule 1: Given radius and chord, find the distance — Take the half chord and use d = √(r² − half chord²).

  • Easy: r = 5, chord 8: half 4, d = √(25 − 16) = 3.
  • Medium: r = 13, chord 24: half 12, d = √(169 − 144) = 5.
  • Exam type: r = 17, chord 30: half 15, d = √(289 − 225) = √64 = 8.

Rule 2: Given radius and distance, find the chord — Half chord = √(r² − d²) and chord = 2 × half chord.

  • Easy: r = 10, d = 6: half √(100 − 36) = 8, chord 16.
  • Medium: r = 20, d = 12: half √(400 − 144) = 16, chord 32.
  • Exam type: r = 25, d = 24: half √(625 − 576) = 7, chord 14. The farther the chord from the centre, the shorter it is.

Rule 3: Two parallel chords: two distances — Find the distances d₁ and d₂ of the two parallel chords from the centre. If the chords are on the same side of the centre, the gap is d₁ − d₂; on opposite sides it is d₁ + d₂.

  • Easy: r = 10, chords 16 and 12: distances 6 and 8.
  • Medium: If on the same side, the gap between the chords is 8 − 6 = 2.
  • Exam type: If on opposite sides the gap is 8 + 6 = 14. (14 is less than the diameter 20, so it is possible.)
Remember

r² = d² + (half chord)². This one line answers every question.

Common mistake

Using the full chord in the square. The half chord must be used.

Shortcut

Watch for right-triangle triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25. Then no square root needs working out. Memory clue: r² = d² + (c ÷ 2)²: "the radius is the hypotenuse".

Spot the difference

GivenTo findFormula
r and chorddistance dd = √(r² − half²)
r and dchordchord = 2√(r² − d²)
d and chordradius rr = √(d² + half²)

Practice questions

Easy
  1. 1If r = 5 and the chord is 6, find the distance.
  2. 2If r = 10 and d = 8, find the chord.
  3. 3If the half chord is 12 and d = 5, find the radius.
Medium
  1. 4If r = 17 and the chord is 16, find the distance.
  2. 5The diameter is 26 and the chord is 24. Find the distance.
  3. 6The chord is 48 and the distance from the centre is 7. Find the radius.
Challenging
  1. 7In a circle r = 10 two parallel chords of 12 and 16 lie on opposite sides of the centre. How far apart are they?
  2. 8If the same chords lie on the same side of the centre, how far apart are they?
Show answers (try them yourself first)
  1. 1Half 3 ⇒ d = √(25 − 9) = 4. 2) Half √(100 − 64) = 6 ⇒ chord 12. 3) r = √(25 + 144) = 13.
  2. 4Half 8 ⇒ d = √(289 − 64) = 15. 5) r = 13, half 12 ⇒ d = √(169 − 144) = 5. 6) Half 24 ⇒ r = √(49 + 576) = 25.
  3. 7Chord 12: d = √(100 − 36) = 8. Chord 16: d = √(100 − 64) = 6. On opposite sides: 8 + 6 = 14. 8) On the same side: 8 − 6 = 2.

The incircle

From daily life

We want to cut the largest possible round piece from inside a triangular tile. It touches all three edges and does not go outside. That is the incircle of the triangle.

Simple definition

The circle inside a triangle touching all three sides is the incircle. Its centre is the incentre (the meeting point of the three angle bisectors) and its radius is the inradius. The incentre is always inside the triangle and equally far from all three sides.

Live figure

The incircle

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Incircle and inradius of a right triangleIncircle: the circle touching all three sidesABCI8610Incentre I = meetingpoint of the anglebisectors, insideRight triangle 6, 8, 10:r = (6 + 8 − 10) ÷ 2r = 2(area 24 ÷ semi-perimeter 12 = 2)

In the figure: The incircle of the right triangle 6, 8, 10: inradius 2.

Rule 1: Incentre = meeting point of angle bisectors — The bisectors of two angles meet at the incentre. The bisector of the third angle also passes through the same point.

  • Easy: The bisector of ∠B = 80° divides it into 40° and 40°.
  • Medium: In ∆ABC, ∠A = 60°, ∠B = 80° and I is the incentre. ∠IAB = 30°, ∠IBA = 40° ⇒ ∠AIB = 180 − 70 = 110°.
  • Exam type: The incentre is always inside; in an equilateral triangle it is also where the three altitudes meet.

Rule 2: Inradius of a right triangle: r = (a + b − c) ÷ 2 — a and b are the sides forming the right angle and c is the hypotenuse. Area = ½ × r × perimeter also holds: r = area ÷ semi-perimeter.

  • Easy: 3, 4, 5: r = (3 + 4 − 5) ÷ 2 = 1. Check: area 6, semi-perimeter 6, 6 ÷ 6 = 1.
  • Medium: 6, 8, 10: r = (6 + 8 − 10) ÷ 2 = 2. Check: area 24, semi-perimeter 12, 24 ÷ 12 = 2.
  • Exam type: 5, 12, 13: r = (5 + 12 − 13) ÷ 2 = 2 (area 30 ÷ semi-perimeter 15 = 2). 8, 15, 17: r = (8 + 15 − 17) ÷ 2 = 3.

Rule 3: Other triangles: r = area ÷ semi-perimeter — Find the area (base × height ÷ 2) and the semi-perimeter s = (a + b + c) ÷ 2, then divide.

  • Easy: Sides 13, 14, 15: s = 21, area 84 (Heron), r = 84 ÷ 21 = 4.
  • Medium: Isosceles 10, 10, 12: height √(100 − 36) = 8, area ½ × 12 × 8 = 48, s = 16, r = 48 ÷ 16 = 3.
  • Exam type: The area of the incircle (π = 22/7) with r = 3 is 22/7 × 9 = 198/7 ≈ 28.29.
Remember

The incentre is always inside the triangle. It is the meeting point of angle bisectors, not of perpendicular bisectors.

Common mistake

Mixing up incentre and circumcentre. The incircle touches the sides; the circumcircle passes through the vertices.

Shortcut

For a right triangle: add the two short sides, subtract the hypotenuse, divide by 2. Memory clue: The incircle touches all three sides from inside; r = area ÷ semi-perimeter.

Spot the difference

IncircleCircumcircle
Touches / passestouches all three sidespasses through all three vertices
Centremeeting point of angle bisectorsmeeting point of perpendicular bisectors
Position of centrealways insideinside, on the hypotenuse or outside

Practice questions

Easy
  1. 1What does the incircle touch?
  2. 2Into which angles does the bisector of ∠B = 70° divide it?
  3. 3What is the inradius of a triangle with sides 3, 4, 5?
Medium
  1. 4Find the inradius of the triangle with sides 5, 12, 13.
  2. 5Find the inradius of the triangle with sides 8, 15, 17.
  3. 6In ∆ABC, ∠A = 50°, ∠B = 70° and I is the incentre. Find ∠AIB.
Challenging
  1. 7Find the inradius of the isosceles triangle with sides 10, 10, 12.
  2. 8Find the inradius of the triangle with sides 13, 14, 15 (area 84).
Show answers (try them yourself first)
  1. 1All three sides. 2) 35° and 35°. 3) (3 + 4 − 5) ÷ 2 = 1.
  2. 4(5 + 12 − 13) ÷ 2 = 2. 5) (8 + 15 − 17) ÷ 2 = 3. 6) ∠IAB = 25°, ∠IBA = 35° ⇒ ∠AIB = 180 − 60 = 120°.
  3. 7Height 8, area 48, s = 16, r = 3. 8) s = 21, r = 84 ÷ 21 = 4.

The circumcircle

From daily life

Three friends stand at three places in a field and a flag must be placed at the same distance from all three. Once that point is found, a circle through all three can be drawn around it. That is the circumcircle.

Simple definition

The circle through all three vertices of a triangle is the circumcircle. Its centre is the circumcentre (the meeting point of the perpendicular bisectors of the three sides) and its radius is the circumradius. The circumcentre is equally far from all three vertices.

Live figure

The circumcircle

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Circumcircle of a right triangle: circumradius is half the hypotenuseCircumcircle: the circle through all three verticesABCOCircumcentre O = meetingpoint of theperpendicular bisectorsRight triangle 6, 8, 10:O = midpoint of ACR = 10 ÷ 2 = 5

In the figure: In a right triangle the circumcentre is at the midpoint of the hypotenuse; the circumradius is half the hypotenuse.

Rule 1: Circumcentre = meeting point of perpendicular bisectors — Draw the perpendicular bisectors of two sides; where they meet is the circumcentre. Its distance from every vertex is equal.

  • Easy: The perpendicular bisector of AB shows all the points equidistant from A and B.
  • Medium: For the circumcentre O we have OA = OB = OC (all radii).
  • Exam type: Position of the circumcentre: inside for an acute triangle, on the hypotenuse for a right triangle, outside for an obtuse triangle.

Rule 2: Circumradius of a right triangle: R = hypotenuse ÷ 2 — The midpoint of the hypotenuse is the circumcentre. So the hypotenuse gives R.

  • Easy: 3, 4, 5: R = 5 ÷ 2 = 2.5.
  • Medium: 6, 8, 10: R = 10 ÷ 2 = 5 (figure).
  • Exam type: 5, 12, 13: R = 6.5. 9, 12, 15: hypotenuse 15 ⇒ R = 7.5 and r = (9 + 12 − 15) ÷ 2 = 3.

Rule 3: Other triangles: R = abc ÷ (4 × area) — Multiply the three sides and divide by four times the area.

  • Easy: 3, 4, 5: abc = 60, area 6 ⇒ R = 60 ÷ 24 = 2.5 (matches the answer above).
  • Medium: Isosceles 10, 10, 12: abc = 1200, area 48, 4 × 48 = 192, R = 1200 ÷ 192 = 6.25.
  • Exam type: 13, 14, 15 (area 84): abc = 2730, 4 × 84 = 336, R = 2730 ÷ 336 = 8.125.
Remember

The circumcircle passes through the vertices. Its centre is where the perpendicular bisectors of the sides meet. In a right triangle R = hypotenuse ÷ 2.

Common mistake

Using the inradius formula for the circumradius. r (in-) and R (circum-) differ: for 6-8-10, r = 2 and R = 5.

Shortcut

Check: R is always greater than r (R ≥ 2r). Memory clue: The circumcircle passes through the vertices; in a right triangle R = hypotenuse ÷ 2.

Spot the difference

TriangleCircumcentreR
Acuteinside the triangleabc ÷ 4A
Rightmidpoint of the hypotenusehypotenuse ÷ 2
Obtuseoutside the triangleabc ÷ 4A

Practice questions

Easy
  1. 1Through what does the circumcircle of a triangle pass?
  2. 2The circumcentre is the meeting point of what?
  3. 3If the hypotenuse of a right triangle is 10, what is the circumradius?
Medium
  1. 4Find the circumradius of the triangle with sides 5, 12, 13.
  2. 5Find the circumradius of the triangle with sides 8, 15, 17.
  3. 6Where is the circumcentre of an obtuse triangle?
Challenging
  1. 7Find the circumradius of the triangle with sides 10, 10, 12 (area 48).
  2. 8The sides forming the right angle in a right triangle are 9 and 12. Find r and R.
Show answers (try them yourself first)
  1. 1Through all three vertices. 2) The perpendicular bisectors of the three sides. 3) 10 ÷ 2 = 5.
  2. 413 ÷ 2 = 6.5. 5) 17 ÷ 2 = 8.5. 6) Outside the triangle.
  3. 7R = 1200 ÷ 192 = 6.25. 8) Hypotenuse 15. r = (9 + 12 − 15) ÷ 2 = 3 and R = 7.5.