Basic properties of circles Examples II


  • Examples
    In the figure, O is the centre of the circle. AOC is a straight line. If AB and BC are tangents to the circle such that \(AB = 10\) and \(BC = 15\),
    [FigureMissing Latex_Basic properties of circles 02 Q47
    (a) find the radius of the circle.
    (b) find the area of the shaded region. (Express your answer in terms of π.)
  • Solutions
    (a) Suppose AB and BC touch the circle at P and Q respectively.
    [FigureMissing Latex_Basic properties of circles 02 Q47
    \(\angle OQC = 90^\circ \) (tangent ⊥ radius)
    \(\angle OPA = 90^\circ \) (tangent ⊥ radius)
    ∴ OPBQ is a square.
    Obviously, \(\Delta OCQ \sim \Delta ACB\).
    ∴ \(\frac{{OQ}}{{AB}} = \frac{{CQ}}{{CB}}\) (corr. sides, ∼ Δs)
    \(\)\(\begin{array}{1}\frac{{OQ}}{{10}} = \frac{{15 - OP}}{{15}}\\\frac{{OQ}}{2} = \frac{{15 - OQ}}{3}\\3OQ = 30 - 2OQ\\5OQ = 30\\OQ = 6\end{array}\)
    ∴ The radius of the circle is 6.

    (b) Area of the shaded region\(\begin{array}{l} = \frac{1}{2} \times 10 \times 15 - \frac{1}{2}{(6)^2}\pi - [{6^2} - \frac{1}{4}{(6)^2}\pi ]\\
    = {\underline{\underline {39 - 9\pi }} ^{}}\end{array}\)


  • Examples
    In the figure, AB is the tangent to the larger circle at B. The circumferences of the larger circle and the smaller circle intersect at C and D. P and Q are points on the smaller circle such that DPB and CQB are straight lines. Prove that AB // PQ.
    [FigureMissing Latex_Basic properties of circles 02 Q66
  • Solutions
    \(\angle DCQ = \angle QPB\) (ext. Δ, cyclic quad.)
    \(\angle ABP = \angle DCQ\) (Δ in alt. segment)
    ∴ \(\angle ABP = \angle QPB\)
    ∴ AB // PQ (alt. Δs eq.)


  • Examples
    In the figure, BC is the tangent to the circle at C. Chord AD is produced to meet BC at B. \(CB = CA\).
    [FigureMissing Latex_Basic properties of circles 02 Q69
    (a) Prove that \(\Delta ABC \sim \Delta CBD\).
    (b) If \(CA = 12\) and \(CD = 7.5\), find the length of AD.
  • Solutions
    (a) In ΔABC and ΔCBD,
    \(\angle ABC = \angle CBD\) (common side)
    \(\angle BAC = \angle BCD\) (∠ in alt. segment)
    \(\angle ACB + \angle ABC + \angle BAC = 180^\circ \) (∠ sum of Δ)
    \(\angle ACB = 180^\circ - \angle ABC - \angle BAC\)
    \(\angle CDB + \angle CBD + \angle BCD = 180^\circ \) (∠ sum of Δ)
    \(\begin{array}{1}\angle CDB = 180^\circ - \angle CBD - \angle BCD\\ = 180^\circ - \angle ABC - \angle BAC\end{array}\) (proved)
    ∴ \(\angle ACB = \angle CDB\)
    ∴ \(\Delta ABC \sim \Delta CBD\) (equiangular)

    (b) \(\begin{array}{1}CB = CA\\ = 12\end{array}\) (given)
    ∵ \(\Delta ABC \sim \Delta CBD\) (proved)
    ∴ \(\begin{array}{1}\frac{{AC}}{{CD}} = \frac{{BC}}{{BD}}\\\frac{{12}}{{7.5}} = \frac{{12}}{{BD}}\\BD = 7.5\end{array}\) (corr. sides, ∼ Δs)
    ∴ \(\begin{array}{1}\frac{{AC}}{{CD}} = \frac{{AB}}{{CB}}\\\frac{{12}}{{7.5}} = \frac{{7.5 + AD}}{{12}}\\7.5 + AD = 19.2\\AD = \underline{\underline {11.7}} \end{array}\) (corr. sides, ∼ Δs)