Chapter 10: Circles — notes

1. 1. First, recall: the parts of a circle

   Take your compass and draw a circle in your notebook.

   • • Circle = all points in a plane at the same distance (radius) from a fixed point (centre O)
   • • Chord: a line segment joining two points of the circle
   • • Diameter: a chord through the centre
   • • Diameter = 2 × radius
   • • The perpendicular from the centre to a chord bisects the chord
   • • OM ⊥ AB ⇒ AM = MB
   • • Distance of the chord from the centre = OM
   • • Example: OA = 13 cm, AB = 24 cm, OM = ?
   • • AM = AB/2 = 24/2 = 12 cm
   • • OM² = OA² − AM²
   • • OM² = 13² − 12² = 169 − 144 = 25
   • • OM = √25 = 5 cm

2. 2. A circle and a line: three situations

   The figure, a circle and a line PQ near it.

   • • Figure: a circle and a line PQ in the same plane
   • • Situation 1: no common point ⇒ non-intersecting line
   • • Situation 2: two common points A and B ⇒ secant (AB is a chord)
   • • Situation 3: exactly one common point A ⇒ tangent
   • • Common points of a line and a circle: 0, 1 or 2, never more
   • • Rope over a pulley: a tangent to the circle of the pulley

3. 3. A tangent: a special case of a secant

   Start turning the straight wire.

   • • Figure: a straight wire pivoted at P on a circular wire, turned slowly
   • • Most positions: two common points ⇒ secant
   • • Exactly one position: only P common ⇒ tangent at P
   • • At a point of a circle there is one and only one tangent
   • • Common point of the tangent and the circle = point of contact
   • • Figure: secant PQ cutting the circle at A and B, moved parallel to itself
   • • The chord AB gets shorter as the line moves away from the centre
   • • A and B coincide ⇒ the secant becomes a tangent
   • • For each direction: exactly two parallel tangents
   • • They touch the circle at the two ends of a diameter

4. 4. Theorem 10.1: the tangent is perpendicular to the radius

   The first big theorem of this chapter.

   • • Figure: circle with centre O; tangent XY at the point P; radius OP
   • • Theorem 10.1: the tangent at any point of a circle is perpendicular to the radius through the point of contact
   • • Given: a circle with centre O and a tangent XY at the point P
   • • To prove: OP ⊥ XY
   • • Construction: take any point Q on XY other than P, and join OQ
   • • Proof: every point of the tangent XY, other than P, lies outside the circle
   • • ∴ Q lies outside the circle
   • • OQ > OP (Q is outside the circle, OP is a radius)
   • • For every point Q of XY (Q ≠ P): OQ > OP
   • • ∴ OP is the shortest distance from O to the line XY
   • • The shortest segment from a point to a line is the perpendicular ⇒ OP ⊥ XY
   • • Hence proved.
   • • At a point of a circle: one and only one tangent
   • • The line of the radius through the point of contact = the normal at that point
   • • Angle between a tangent and the radius at the point of contact = 90°
   • • Example: OP = 5 cm, OQ = 13 cm, PQ = ?
   • • ∠OPQ = 90° (Theorem 10.1)
   • • In right △OPQ, hypotenuse = OQ
   • • OQ² = OP² + PQ²
   • • PQ² = OQ² − OP² = 13² − 5² = 169 − 25 = 144
   • • 12 × 12 = 144
   • • PQ = √144 = 12 cm

5. 5. How many tangents from a point

   It depends on where the point sits.

   • • Figure: a circle with a point P inside it, on it, and outside it
   • • Case 1: P inside the circle ⇒ no tangent
   • • Case 2: P on the circle ⇒ exactly one tangent
   • • Case 3: P outside the circle ⇒ exactly two tangents PT₁ and PT₂
   • • Length of the tangent from P = PT₁ (from P to the point of contact)
   • • Example: PQ = 24 cm, OQ = 25 cm, radius OP = ?
   • • ∠OPQ = 90° (Theorem 10.1) ⇒ OQ is the hypotenuse
   • • OP² = OQ² − PQ²
   • • OP² = 25² − 24² = 625 − 576 = 49
   • • OP = √49 = 7 cm

6. 6. Theorem 10.2: the tangents from an external point are equal

   The second big theorem.

   • • Figure: centre O; external point P; tangents PQ and PR touching at Q and R; OP, OQ, OR joined
   • • Theorem 10.2: the lengths of tangents drawn from an external point to a circle are equal
   • • Given: a circle with centre O; PQ and PR are tangents from the external point P
   • • To prove: PQ = PR
   • • Construction: join OP, OQ and OR
   • • Proof: ∠OQP = 90° (Theorem 10.1, OQ ⊥ PQ)
   • • ∠ORP = 90° (Theorem 10.1, OR ⊥ PR)
   • • In △OQP and △ORP:
   • • OP = OP (common hypotenuse)
   • • OQ = OR (radii of the same circle)
   • • ∴ △OQP ≅ △ORP (RHS)
   • • ∴ PQ = PR (CPCT)
   • • Hence proved.
   • • PQ² = OP² − OQ², PR² = OP² − OR²
   • • OQ = OR ⇒ PQ² = PR² ⇒ PQ = PR
   • • △OQP ≅ △ORP ⇒ ∠OPQ = ∠OPR (CPCT)
   • • OP bisects ∠QPR: the centre lies on the bisector of the angle between the tangents
   • • ∠POQ = ∠POR (CPCT): equal angles at the centre

7. 7. The angle between the two tangents

   A result that comes in the board exam almost every year.

   • • Figure: tangents PA and PB from P; radii OA and OB; quadrilateral OAPB
   • • ∠OAP + ∠APB + ∠OBP + ∠AOB = 360° (angle sum of a quadrilateral)
   • • ∠OAP = ∠OBP = 90° (Theorem 10.1)
   • • 90° + ∠APB + 90° + ∠AOB = 360°
   • • ∠APB + ∠AOB = 360° − 180° = 180°
   • • ∠APB + ∠AOB = 180°: the angle between the tangents is supplementary to the angle at the centre
   • • Example: ∠AOB = 120° ⇒ ∠APB = 180° − 120° = 60°
   • • Example: tangents TP and TQ from T; prove ∠PTQ = 2∠OPQ
   • • Let ∠PTQ = θ
   • • TP = TQ (Theorem 10.2) ⇒ ∠TPQ = ∠TQP
   • • ∠TPQ + ∠TQP = 180° − θ
   • • ∠TPQ = 1/2(180° − θ) = 90° − θ/2
   • • ∠OPT = 90° (Theorem 10.1)
   • • ∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − θ/2) = θ/2
   • • ∴ ∠PTQ = θ = 2∠OPQ
   • • Hence proved.

8. 8. Concentric circles and a chord

   Circles with the same centre are called concentric circles.

   • • Figure: concentric circles with centre O; chord AB of the larger circle touching the smaller circle at P
   • • For the smaller circle: AB is a tangent at P
   • • For the larger circle: AB is a chord
   • • Example 1: prove that P bisects AB
   • • OP ⊥ AB (Theorem 10.1: AB is a tangent to the smaller circle at P)
   • • AB is a chord of the larger circle and OP ⊥ AB
   • • ∴ AP = PB (the perpendicular from the centre bisects the chord)
   • • Hence proved.
   • • Example: OA = 5 cm, OP = 3 cm, AB = ?
   • • AP² = OA² − OP² = 5² − 3² = 25 − 9 = 16
   • • AP = √16 = 4 cm
   • • AB = 2 × AP = 2 × 4 = 8 cm

9. 9. Tangents at the ends of a chord

   A slightly longer example, so go step by step with me.

   • • Example: radius OP = 5 cm, chord PQ = 8 cm; the tangents at P and Q meet at T; TP = ?
   • • Figure: OT joined, cutting PQ at R
   • • TP = TQ (Theorem 10.2), ∠PTR = ∠QTR (OT bisects ∠PTQ)
   • • △TPR ≅ △TQR (SAS) ⇒ PR = RQ, ∠TRP = ∠TRQ = 90°
   • • PR = PQ/2 = 8/2 = 4 cm
   • • OR² = OP² − PR² = 5² − 4² = 25 − 16 = 9
   • • OR = √9 = 3 cm
   • • ∠TPR + ∠RPO = 90° (∠TPO = 90°, Theorem 10.1)
   • • ∠RPO + ∠POR = 90° (in △PRO, ∠PRO = 90°)
   • • ⇒ ∠TPR = ∠POR
   • • △TRP ~ △PRO (AA similarity)
   • • TP/PO = RP/RO
   • • TP/5 = 4/3
   • • TP = (4 × 5)/3 = 20/3 cm

10. 10. A circle inside a triangle or a quadrilateral

    The figure, a △ABC, and a circle inside it that touches all three sides.

    • • Figure: △ABC with a circle inside touching BC at P, CA at Q, AB at R
    • • From A: AR and AQ, from B: BR and BP, from C: CP and CQ
    • • AR = AQ, BR = BP, CP = CQ (Theorem 10.2)
    • • Example: AB = 12 cm, AQ = 7 cm, CQ = 4 cm, BC = ?
    • • AR = AQ = 7 cm, BR = AB − AR = 12 − 7 = 5 cm
    • • BP = BR = 5 cm, CP = CQ = 4 cm
    • • BC = BP + PC = 5 + 4 = 9 cm
    • • Figure: quadrilateral ABCD round a circle, touching AB, BC, CD, DA at P, Q, R, S
    • • To prove: AB + CD = AD + BC
    • • AP = AS, BP = BQ, CR = CQ, DR = DS (Theorem 10.2)
    • • AB + CD = (AP + PB) + (CR + RD)
    • • = (AS + BQ) + (CQ + DS)
    • • = (AS + DS) + (BQ + CQ)
    • • = AD + BC
    • • Hence proved.
    • • Quadrilateral round a circle: AB + CD = AD + BC
    • • Parallelogram round a circle ⇒ 2AB = 2AD ⇒ rhombus

AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 10. Written from the textbook, checked twice.