Basic properties of circles Examples II

ExamplesIn 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\(\begin{array}{l} = \frac{1}{2} \times 10 \times 15  \frac{1}{2}{(6)^2}\pi  [{6^2}  \frac{1}{4}{(6)^2}\pi ]\\
= {\underline{\underline {39  9\pi }} ^{}}\end{array}\)

ExamplesIn 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.
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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.)

ExamplesIn the figure, BC is the tangent to the circle at C. Chord AD is produced to meet BC at B. \(CB = CA\).
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(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)