Basic properties of circles - (10.1)

Understand the properties of chords and arcs of a circle

  • Theory


    The chords of equal arcs are equal.

    Inversely,

    Equal chords cut off equal arcs
  • Examples
  • Solutions
  • Graph
  • Theory


    the perpendicular from the centre to a chord bisects the chord
  • Examples
    In the figure, O is the centre of the circle. C is a point on chord AB
    such that OC ⊥ AB. If \(AC = (x - 10){\rm{ cm}}\) and \(BC = \frac{x}{3}{\rm{ cm}}\),
    find the length of AB.
  • Solutions
    \(\because\) OC ⊥ AB (given)
    \(\begin{array}{1}\therefore {\rm{ }}AC =BC\\x - 10 = \frac{x}{3}\\3x - 30 = x\\2x = 30\\x = 15\end{array}\) (line from centre ⊥ chord bisects chord)

    \(\begin{array}{1}\therefore {\rm{ }}AB = AC + BC\\ = [(15 - 10) + \frac{{15}}{3}]{\rm{ cm}}\\ =\underline{\underline {{\rm{10 cm}}}} \end{array}\)
  • Graph


  • Theory
    the straight line joining the centre and the mid-point of a chord which is not a diameter is perpendicular to the chord
    line joining centre to mid-pt. of chord ⊥ chord
  • Examples
    In the figure, O is the centre of the circle. Radius OD intersects chord AB
    at C such that \(AC = CB = 6\). If \(CD = 3\), find the length of OC.
  • Solutions
    \(\angle OCA = 90^\circ \) (line joining centre to mid-pt. of chord ⊥ chord)
    \(\begin{array}{c}OA = OD (radii) \\ OA = OC + 3 \end{array}\)

    In ∆OAC,
    \(\because O{A^2} =O{C^2} + A{C^2}\) (Pyth. theorem)
    \(\begin{array}{1}\therefore {\rm{ }}{(OC + 3)^2} = O{C^2} + {6^2}\\O{C^2} + 6OC + 9 = O{C^2} + 36\\6OC = 27\\OC = \underline{\underline {4.5}} \end{array}\)
  • Graph
  • Theory
    the perpendicular bisector of a chord passes through the centre
  • Examples
    In the figure, AP and BP are equal chords. Chord PQ is perpendicular to chord AB at M.
    Is PQ is a diameter of the circle?
  • Solutions
    In ∆APM and ∆BPM,
    \(AP = BP\) (given)
    \(PM = PM\) (common side)
    \(\angle AMP = \angle BMP = 90^\circ \) (given)
    \(\Delta APM \cong \Delta BPM\) (R.H.S.)

    \(\because\ AM = BM\) (corr. sides, ≅s)
    \(\therefore\) PQ passes through the centre. ( ⊥bisector of chord passes through centre)
    \(\therefore\) PQ is a diameter.
  • Graph
  • Theory
    equal chords are equidistant from the centre
  • Examples
    In the figure, O is the centre of the circle. M and N are points on chords AB and AC
    respectively such that \(OM \bot AB\) and \(ON \bot AC\). \(AB = 4\), \(AN = 2\), \(OM = 3\).
    Find the length of ON.
  • Solutions
    \(ON \bot AC\) (given)
    \(AN = CN\) (line from centre ⊥ chord bisects chord)
    \(\begin{array}{c}AC = 2AN\\ = 2 \times {2^{}}\\ = {4^{}}\end{array}\)
    \(AC = AB = 4\)
    \(\begin{array}{1}ON = OM\\ ON = {3^{}}\end{array}\) (eq. chords equidistant from centre)
  • Graph
  • Theory
    chords equidistant from the centre are equal
  • Examples
    In the figure, O is the centre of the circle. A and B are points on chords
    MN and HK respectively such that \(OA \bot MN\) and \(OB \bot HK\).
    If \(OA = OB\) and \(AM = 3{\rm{ cm}}\), then \(HK = \)
  • Solutions
    \(OA \bot MN\) (given)
    \(AN = AM\) (⊥ from centre bisects chord)
    \(\begin{array}{1}MN = 2AM\\ = 2 \times {\rm{3 c}}{{\rm{m}}^{}}\\ = {\rm{6 c}}{{\rm{m}}^{}}\end{array}\)
    \(OA = OB\) (given)
    \(\begin{array}{1}HK = MN\\ = 6{\rm{ c}}{{\rm{m}}^{}}\end{array}\) (chords equidistant from centre eq.)