### Basic properties of circles Examples II

• 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.
• (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$$\begin{array}{l} = \frac{1}{2} \times 10 \times 15 - \frac{1}{2}{(6)^2}\pi - [{6^2} - \frac{1}{4}{(6)^2}\pi ]\\ = {\underline{\underline {39 - 9\pi }} ^{}}\end{array}$$
• 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
• $$\angle DCQ = \angle QPB$$ (ext. Δ, cyclic quad.)
$$\angle ABP = \angle DCQ$$ (Δ in alt. segment)
∴ $$\angle ABP = \angle QPB$$
∴ AB // PQ (alt. Δs eq.)
• 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.
• (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)