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 
ExamplesIn the figure, O is the centre of the circle. AD is a diameter. If \(AB = BC = CD\), find x.

Solutions\(\because\) \(AB = BC = CD\) (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)

Theory
the perpendicular from the centre to a chord bisects the chord

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

Theory
the straight line joining the centre and the midpoint of a chord ⊥ to the chord 
ExamplesIn 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 midpt. 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)
\(\begin{array}{1}\therefore {\rm{ }}{(OC + 3)^2} = O{C^2} + {6^2}\\O{C^2} + 6OC + 9 = O{C^2} + 36\\6OC = 27\\OC = \underline{\underline {4.5}} \end{array}\)

Theory
the perpendicular bisector of a chord passes through the centre 
ExamplesIn 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? 
SolutionsIn ∆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.

Theory
equal chords are equidistant from the centre 
ExamplesIn 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

Theorychords equidistant from the centre are equal

ExamplesIn 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.)