### Basic properties of circles - (10.1)

#### Understand the properties of chords and arcs of a circle

• The chords of equal arcs are equal. Inversely, Equal chords cut off equal arcs
• In the figure, O is the centre of the circle. AD is a diameter. If $$AB = BC = CD$$, find x.

• $$\because$$ $$AB = BC = CD$$ (given)
$$\therefore$$ $$\angle AOB = \angle BOC = \angle COD = x$$ (eq. chords, eq. $$\angle$$s)
$$\begin{array}{1}\angle AOB + \angle BOC + \angle COD = 180^\circ \\x + x + x = 180^\circ \\3x = 180^\circ \\x = \underline{\underline {60^\circ }} \end{array}$$ (adj. $$\angle$$s on st. line)

• the perpendicular from the centre to a chord bisects the chord
• 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.

• $$\because$$ 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)

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

• the straight line joining the centre and the mid-point of a chord ⊥ to the chord
• 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.
• $$\angle OCA = 90^\circ$$ (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)
$$\begin{array}{1}\therefore {\rm{ }}{(OC + 3)^2} = O{C^2} + {6^2}\\O{C^2} + 6OC + 9 = O{C^2} + 36\\6OC = 27\\OC = \underline{\underline {4.5}} \end{array}$$

• the perpendicular bisector of a chord passes through the centre
• 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?
• 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.

• equal chords are equidistant from the centre
• 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.
• $$ON \bot AC$$ (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)
• chords equidistant from the centre are equal
• 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 $$HK =$$
• $$OA \bot MN$$ (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.)