Chapter 9

Circles

  • A circle is the set of all points that are at fixed distance from a fixed point.
  • အမှတ်သေတစ်ခုမှသတ်မှတ်ထားသောအကွာအဝေးရှိသောအမှတ်များအားလုံး၏အစုကိုစက်ဝိုင်းဟုခေါ်သည်။
  • The fixed point producing a circle is called the centre of the circle.
  • ထိုအမှတ်သေကိုစက်ဝိုင်း၏ဗဟိုဟုခေါ်သည်။
  • A circle with centre O is called circle O and denoted by ⊙ O.
  • Oဗဟိုရှိသောစက်ဝိုင်းကို circle O ဟုခေါ်ပြီးသင်္ကေတ ⊙ Oဟုရေးလေ့ရှိသည်။
  • Circles having the same centre are called concenteric circles.
  • ဗဟိုတူသောစက်ဝိုင်းများကိုဗဟိုတူစက်ဝိုင်းများဟုခေါ်သည်။
  • A radius is a segment joining the centre and a point on the circle.
  • စက်ဝိုင်းပေါ်ရှိအမှတ်တစ်ခုနှင့်ဗဟိုတို့ဆက်သွယ်သောမျဉ်းပြတ်ကိုစက်ဝိုင်း၏အချင်းဝက်ဟုခေါ်သည်။
  • A segment joining two points on a circle is called a chord of the circle.
  • စက်ဝိုင်းပေါ်ရှိအမှတ်နှစ်ခုကိုဆက်သွယ်သောမျဉ်းပြတ်တစ်ခုကိုလေးကြိုးဟုခေါ်သည်။
  • A diameter is a chord passing through the centre of a circle.
  • စက်ဝိုင်းတစ်ခု၏ဗဟိုကိုဖြတ်သွားသောလေးကြိုးကိုအချင်းဟုခေါ်၏။
  • In a circle, a diameter is a longest chord.
  • စက်ဝိုင်းတစ်ခုတွင်အချင်းသည်အရှည်ဆုံးလေးကြိုးတစ်ခုဖြစ်သည်။
  • Circles having the same radius are called congruent circles.
  • အချင်းဝက်တူသောစက်ဝိုင်းများကိုထပ်တူညီစက်ဝိုင်းများဟုခေါ်သည်။
  • A secant of a circle is a line that intersects the circle at two points.
  • စက်ဝိုင်းတစ်ခုပေါ်ရှိအမှတ်နှစ်ခုကိုဖြတ်သွားသောမျဉ်းဖြောင့်ကိုဝန်းဖြတ်မျဉ်းဟုခေါ်သည်။
  • A line touching the circle at one point only is called a tangent to the circle.
  • စက်ဝိုင်းပေါ်ရှိအမှတ်တစ်ခုကိုထိသည့်မျဉ်းဖြောင့်ကိုဝန်းထိမျဉ်းဟုခေါ်သည်။
  • This touching point is called the point of contact.
  • ဝန်းထိမျဉ်းထိနေသောအမှတ်ကိုဝန်းထိမှတ်ဟုခေါ်သည်။

  • An arc is a part of a circle.
  • စက်ဝိုင်းပေါ်ရှိအပိုင်းတစ်ခုကိုအဝန်းပိုင်းဟုခေါ်သည်။
  • A semicircle is a half part of a circle.
  • စက်ဝိုင်း၏တစ်ဝက်ကိုစက်ဝိုင်းခြမ်းဟုခေါ်သည်။
  • A minor arc is an arc shorter than a semicircle.
  • စက်ဝိုင်းခြမ်း၏အဝန်းအလျားထက်ငယ်သောအဝန်းပိုင်းကိုအဝန်းပိုင်းငယ်ဟုခေါ်သည်။
  • A major arc is an arc longer than a semicircle.
  • စက်ဝိုင်းခြမ်း၏အဝန်းအလျားထက်ပိုရှည်သောအဝန်းပိုင်းကိုအဝန်းပိုင်းကြီးဟုခေါ်သည်။

9.1 Angles in a Circle

Centeral Angles and Inscribed Angles

A central angle is an angle substended by an arc (or chord) of a circle at the centre.
စက်ဝိုင်းတစ်ခု၏အဝန်းပိုင်းတစ်ခု(သို့မဟုတ်လေးကြိုးတစ်ခု)ကယင်းစက်ဝိုင်း၏ဗဟို၌ခံဆောင်ထားသောထောင့်ကိုဗဟိုခံထောင့်ဟုခေါ်သည်။

An inscribed angle is an angle substended by an arc (or chord) of a circle at a point on the other arc.
စက်ဝိုင်းတစ်ခု၏အဝန်းပိုင်းတစ်ခု(သို့မဟုတ်လေးကြိုးတစ်ခု)ကကျန်အဝန်းပိုင်းပေါ်ရှိအမှတ်တစ်ခု၌ခံဆောင်ထားသောထောင့်ကိုအဝန်းခံထောင့်ဟုခေါ်သည်။


∠ COD is a central angle substended by arc CD (or chord CD).
∠ PQR is an inscribed angle substended by arc PR (or chord PR).
Theorem 1
The central angle is twice the inscribed angle substended by the same arc.
စက်ဝိုင်းတစ်ခုတွင်အဝန်းပိုင်းတစ်ခုကဗဟို၌ခံဆောင်ထားသောထောင့်သည်ယင်းအဝန်းပိုင်းကကျန်စက်ဝန်းပေါ်ရှိအမှတ်တစ်ခု၌ခံဆောင်ထားသောထောင့်၏နှစ်ဆရှိသည်။

In ⊙O, AB substends central angle AOB and inscribed angle ACB.Then α= 2 β.
In ⊙O, arc PSQ substends central angle POQ and inscribed angle PRQ.Then θ= 2 γ.

Corollary 1.1
Inscribed angles substended by the same arc are equal.
အဝန်းပိုင်းတစ်ခုကကျန်အဝန်းပိုင်းပေါ်ရှိအမှတ်များ၌ခံဆောင်ထားသောထောင့်များတူညီကြ၏။

∠ ACB and ∠ ADB are inscribed angles substended by arc AB. Then α= β.

Corollary 1.2
An inscribed angle substended by a diameter is a right angle.
အချင်းမျဉ်းတစ်ခုကစက်ဝိုင်းပေါ်ရှိအမှတ်တစ်ခု၌ခံဆောင်ထားသောထောင့်သည်ထောင့်မှန်တစ်ခုဖြစ်၏။

In ⊙ O, AB is a diameter. Then ∠ ACB=$90^\circ$.


Cyclic Quadrilateral

A quadrilateral whose vertices lie on a circle is called a cyclic quadrilateral.
စတုဂံတစ်ခု၏ထောင့်စွန်းလေးခုစလုံးသည်စက်ဝိုင်းတစ်ခုတည်းပေါ်ကျရောက်နေသောစတုဂံကိုစက်ဝိုင်းတွင်းကျစတုဂံဟုခေါ်၏။

Theorem 2
Opposite angles of a cyclic quadrilateral are supplementry.
စက်ဝိုင်းတွင်းကျစတုဂံတစ်ခု၏မျက်နှာချင်းဆိုင်ထောင့်နှစ်ခုသည်ထောင့်ဖြောင့်ဖြည့်ဖက်များဖြစ်ကြ၏။

ABCD is a cyclic quadrilateral. Then ∠ A+ ∠ C=$180^\circ$ and ∠ B+ ∠ D=$180^\circ$.

Corollary 2.1
The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle of the quadrilateral.
စက်ဝိုင်းတွင်းကျစတုဂံတစ်ခု၏အနားတစ်ဖက်ကိုဆက်ဆွဲ၍ဖြစ်ပေါ်လာသောထောင့်သည်အတွင်းထောင့်၏မျက်နှာချင်းဆိုင်ထောင့်နှင့်တူညီသည်။

ABCD is a cyclic quadrilateral. Then ∠ PBC= ∠ D.

Theorem 3
In the same circle or in congruent circles,
  1. equal arcs substend equal central angles,
  2. arcs substending equal central angles are equal.
စက်ဝိုင်းတစ်ခုတည်းတွင်ဖြစ်စေ၊အရွယ်တူစက်ဝိုင်းများတွင်ဖြစ်စေ၊
  1. တူညီသောအဝန်းပိုင်းများသည်ဗဟို၌တူညီသောထောင့်များကိုခံဆောင်ထား၏။
  2. ဗဟို၌တူညီသောထောင့်များကိုခံဆောင်ထားသည့်အဝန်းပိုင်းများသည်တူကြ၏။

arc AB= arc QR ⇔ ∠ AOB= ∠ QPR.

Theorem 4
In the same circle or in congruent circles, two inscribed angles are equal if and only if the corresponding arc are equal.
စက်ဝိုင်းတစ်ခုတည်းတွင်ဖြစ်စေ၊အရွယ်တူစက်ဝိုင်းများတွင်ဖြစ်စေ၊အဝန်းပေါ်ရှိထောင့်များတူကြလျှင်ယင်းတို့ကိုခံဆောင်ထားသောလိုက်ဖက်အဝန်းပိုင်းများတူကြ၏။အပြန်အလှန်လည်းမှန်၏။


9.2 Properties of Chords

Theorem 5
In the same circle or in congruent circles,
  1. equal chords substend equal central angles
  2. equal central angles cut off equal chords.
စက်ဝိုင်းတစ်ခုတည်းတွင်ဖြစ်စေ၊အရွယ်တူစက်ဝိုင်းများတွင်ဖြစ်စေ၊
  1. အလျားတူသောလေးကြိုးများသည်တူညီသောဗဟိုထောင့်များကိုခံဆောင်ထားပြီး
  2. တူညီသောဗဟိုခံထောင့်များသည်တူညီသောလေးကြိုးများကိုပိုင်းဖြတ်ကြ၏။

⊙P and ⊙Q are congruent.
AB=QR ⇔ θ = β .

Theorem 6
If a line passing through the centre of a circle bisects a chord of the circle, then the line is perpendicular to the chord.
စက်ဝိုင်းတစ်ခု၏ဗဟိုကိုဖြတ်သွားသောမျဉ်းဖြောင့်တစ်ကြောင်းသည်ယင်းစက်ဝိုင်း၏လေးကြိုးတစ်ခုကိုထက်ဝက်ပိုင်းလျှင်ယင်းမျဉ်းဖြောင့်သည်ထိုလေးကြိုးကိုထောင့်မတ်ကျ၏။

l is the line passing through the centre O and bisects chord AB at D. Then l ⊥ AB.

Theorem 7
If a line passing through the centre of a circle is perpendicular to a chord of the circle, then the line bisects the chord.
စက်ဝိုင်းတစ်ခု၏ဗဟိုကိုဖြတ်သွားသောမျဉ်းဖြောင့်တစ်ကြောင်းသည်ယင်းစက်ဝိုင်း၏လေးကြိုးတစ်ခုကိုထောင့်မတ်ကျလျှင်ယင်းမျဉ်းဖြောင့်သည်ယင်းလေးကြိုးကိုထက်ဝက်ပိုင်း၏။

l is the line passing through the centre O and is perpendicular to chord AB at D. Then AD=BD.

Theorem 8
The perpendicular bisector of a chord of a circle passes through the centre.
စက်ဝိုင်းတစ်ခု၏လေးကြိုးတစ်ခုကိုထောင့်မတ်ကျထက်ဝက်ပိုင်းမျဉ်းသည်ယင်းစက်ဝိုင်း၏ဗဟိုကိုဖြတ်သွား၏။

l is the perpendicular bisector of chord AB. Then l passes through the centre O.

Theorem 9
In the same circle or in congruent circles, chords are equal if and only if they are equaidistant from the centre of the circle.
စက်ဝိုင်းတစ်ခုတည်းတွင်ဖြစ်စေ၊အရွယ်တူစက်ဝိုင်းများတွင်ဖြစ်စေ၊အလျားတူသောလေးကြိုးများသည်ယင်းစက်ဝိုင်း၏ဗဟိုမှတူညီစွာကွာဝေးကြပြီး၊ဗဟိုမှတူညီစွာကွာဝေးသောလေးကြိုးများသည်အလျားတူကြ၏။

In ⊙ O, AB=CD ⇔ OE= OF.

Theorem 10
Of any chords of a circle, the greater chord is nearer to the centre and conversely, the chord nearer to the centre is larger.
စက်ဝိုင်းတစ်ခုတွင်လေးကြိုးနှစ်ခုအနက်ပို၍ရှည်သောလေးကြိုးသည်ဗဟိုနှင့်ပို၍နီးပြီးအပြန်အလှန်အားဖြင့်ဗဟိုနှင့်ပိုနီးသောလေးကြိုးသည်အလျားပိုရှည်၏။

In ⊙ O, AB>CD ⇔ OE< OF.

Theorem 11
If two chords of a circle intersect in the circle, the product of the lengths of segments of one chord is equal to the product of the lengths of segments of the other chord.
လေးကြိုးနှစ်ခုတို့သည်စက်ဝိုင်းတွင်းအမှတ်တစ်နေရာ၌ဖြတ်ကြလျှင်လေးကြိုးတစ်ခု၏မျဉ်းပြတ်နှစ်ခုတို့၏အလျားများမြှောက်လဒ်သည်ကျန်လေးကြိုး၏မျဉ်းပြတ်နှစ်ခုတို့၏အလျားများမြှောက်လဒ်နှင့်တူညီသည်။

Chord AB and chord CD intersect at P in the circle.
Then PA.PB=PC.PD

Theorem 12
If two secants are drawn to a circle from an external point, the product of the lengths of one secant and its external segment is equal to the product of the lengths of the other secant and its external segment.
ပြင်ပအမှတ်တစ်ခုမှစက်ဝိုင်းသို့ဝန်းဖြတ်မျဉ်းတစ်ခုဆွဲခဲ့လျှင်မျဉ်းဖြတ်တစ်ခုလုံး၏အလျားနှင့်စက်ဝိုင်းပြင်ပရှိမျဉ်းပြတ်အလျားမြှောက်လဒ်သည် ကျန်မျဉ်းဖြတ်တစ်ခုလုံးနှင့်၎င်း၏စက်ဝိုင်းပြင်ပရှိမျဉ်းပြတ်အလျားမြှောက်လဒ်တို့တူညီကြသည်။

PBA andPDC are two secant segments.
Then PA.PB=PC.PD


Problems

Example 1

In the given figure, O is the centre of the circle, arc PQ= arc QR= arc RS. Find PR.


$\begin{aligned} \operatorname{arcPQ}&=\operatorname{arc} Q R=\operatorname{arc} R S\quad(\text { given }) \\ \therefore \angle P O Q&=\angle Q O R=\angle R O S \quad(\text { Theorem } 3) \\ \text { But } \angle P O Q&+\angle Q O R+\angle R O S=90^{\circ} \\ \therefore \angle P O Q&=\angle Q O R=\angle R O S=\frac{90^{\circ}}{3}=30^{\circ} \\ \angle P O R&=\angle P O Q+\angle Q O R=30^{\circ}+30^{\circ}=60^{\circ} \\ \angle P&=\angle R \quad(\because O R=O P \quad\text { radii }) \\ \therefore \angle P&=\angle R=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ} \\ \therefore \triangle &P O R \text { is equalateral. } \\ \therefore P R&=O R \\ \text { But } O R&=O S \quad(\because \text { radii }) \\ \therefore P R&=O S=3 \mathrm{~cm} \end{aligned}$

Example 2

In ⊙ O, find the values of α, β and θ .

(a)

$\begin{aligned} \angle O A B&=\angle O B A\quad(\because O A= O B \text { radii }) \\ \therefore \angle O A B&=40^{\circ} \\ \alpha&=180^{\circ}-\left(40^{\circ}+40^{\circ}\right)=100^{\circ} \\ \alpha&=2 \beta\quad(\because \text { angles substended by arc } A B) \\ \beta&=\frac{1}{2} \alpha=\frac{1}{2}\left(100^{\circ}\right)=50^{\circ} \\ \theta&=\beta\quad(\because \text { angles substended by } \operatorname{arc} A B) \\ \therefore \theta&=50^{\circ} \\ \end{aligned}$

(b)

$\begin{aligned} \beta&=\dfrac{1}{2} \times 200^{\circ}\quad(\because \text { angles substended by arc } A C B) \\ \beta&=100^{\circ} \\ \alpha+\beta&=180^{\circ}\quad(\because \text { opposite angles of cyclic quadrilateral } A D B C) \\ \alpha&=180^{\circ}-100^{\circ}=80^{\circ} \end{aligned}$

Example 3

In the diagram below, O is the centre of the circle. Find the values of α, β.

(a)

$\begin{aligned} 140^{\circ} &=2 \alpha\quad(\because \text { angles substended by arc } A B C) \\ \alpha &=70^{\circ} \\ \angle B+\alpha&=180^{\circ}\quad(\because \text { opposite angles of cyclic quadrilateral } A B C D)\\ \angle B&= 180^{\circ}-\alpha \\ &=180-70^{\circ} \\ &=110^{\circ} \\ \beta&= \angle B C A\quad(\because \text { arc } A B={ arc } B C) \end{aligned}$
But
$\begin{aligned} \beta_{+} \angle B+\angle B C A &=180^{\circ} \\ 2 \beta+110^{\circ}&=180^{\circ} \\ \beta&=\frac{70^{\circ}}{2}\\ \beta&=35^{\circ} \end{aligned}$

(b)

$\begin{aligned} \angle D C B&=90^{\circ}\quad(\because \text { angle substended by diameter} D B)\\ \angle D C B+\alpha&=100^{\circ} \quad(\because \text { exterior angle of cyclic quadrilateral } A D B C )\\ \alpha&=100^{\circ}-90^{\circ}=10^{\circ} \\ \angle A+40^{\circ}+\alpha&=180^{\circ} \\ \angle A&=180^{\circ}-\left(40^{\circ}+10^{\circ}\right)=130^{\circ}\\ \angle A+\beta&=180^{\circ} \quad(\because \text { opposite angles of cyclic quadrilateral } A D B C )\\ \beta&=180^{\circ}-130^{\circ}=50^{\circ}\\ \end{aligned}$

Example 4

In the given figure, find the values of x and y.


DM is the perpendicular bisector of AB.
Therefore the center is on DM.
PN is the perpendicular bisector of CD.
Therefore the center is also on PN.
DM and PN intersect at Q.
Therefore Q is the centre of the given circle.
Join AQ.
$\begin{aligned} A M &=M B \quad\text { (given) } \\ \therefore A M &=3 \mathrm{~cm} \\ A Q^{2} &=A M^{2}+M Q^{2}=3^{2}+4^{2}=25 \\ A Q &=5 \mathrm{~cm} \\ y &=A Q \quad\text { (radii) } \\ \therefore y &=5 \mathrm{~cm} \\ D N &=N C \quad\text { (given) } \\ \therefore D N &=4 \mathrm{~cm} \\ y^{2} &=x^{2}+D N^{2} \\ x &=\sqrt{y^{2}-D N^{2}} \\ &=\sqrt{5^{2}-4^{2}}\\ &=3 \mathrm{~cm} \end{aligned}$

Example 5

Given ⊙ O, OE= OF, AE= x+2 and CD= 3x-2.Find the radius of the circle.


$\begin{aligned} O E &\perp A B \\ \therefore A B &=2 A E=2(x+2)=2 x+4 \\ \text { since } O E &=O F \text {, then } A B=C D \\ 2 x+4 &=3 x-2 \\ x &=6 \\ \therefore A E &=6+2=8 \\ \end{aligned}$
Join OA.
$\begin{aligned} \text { radius }=O A&=\sqrt{O E^{2}+A E^{2}} \\ &=\sqrt{6^{2}+8^{2}}\\ &=10 \end{aligned}$

Example 6

Given ⊙ O, AE= 12, BE= x and CE= 2x. Find radius of the circle.


$\begin{aligned} O B &\perp C D \\ \therefore E D&=C E=2 x \\ \end{aligned}$
$A B$ and $C D$ intersect at $E$ . \\ $\begin{aligned} \therefore A E &\cdot E B=C E \cdot E D \\ 12 \cdot x&=2 x \cdot 2 x \\ x&=3 \\ A B&=A E+E B=12+x=15 \\ \end{aligned}$
$\begin{aligned} \therefore \text { radius }=O A&=\dfrac{1}{2} A B\\ &=\dfrac{1}{2}(15)\\ &=7.5 \end{aligned}$

Example 7

Given ⊙ O, radius= 6, PC= 3, PA= 5 and EF= 6. Find AB and PE.


$\begin{aligned} &D C=\text { diameter }=12 \\ &P D=P C+C D=3+12=15 \\ &P A \cdot P B=P C \cdot P D \quad(\text { Theorem } 12) \\ &5 \cdot(5+A B)=3 \times 15 \\ &5+A B=9 \\ &A B=4 \\ &P E \cdot P F=P C \cdot P D \quad(\text { Theorem } 12) \\ &P E(P E+6)=3 \times 15 \\ &P E^{2}+6 P E=45 \\ &P E^{2}+6 P E+9=45+9 \\ &(P E+3)^{2}=54 \\ &P E+3=\sqrt{54} \\ \end{aligned}$
$\begin{aligned} \therefore P E&=\sqrt{54}-3\\ &=3 \sqrt{6}-3 \\ \end{aligned}$

Exercise 9.1

  1. The point O is the centre of the given circle. Find the values of x, y and z.


  2. (a)


    $\begin{aligned} x=70^{\circ} &\quad(\because\text { Corollary 1.1 }) \\ \alpha+70^{\circ} &=180^{\circ}\quad(\because\text { Theorem 2) }\\ \alpha &=180^{\circ}-70^{\circ}=110^{\circ} \\ \beta &=30^{\circ} \quad(\because\text { alternative angles) } \\ \alpha+\beta+y &=180^{\circ} \\ y &=180^{\circ}-\left(110^{\circ}+30^{\circ}\right) \\ &=40^{\circ} \\ \end{aligned}$


    (b)


    $\begin{aligned} x=& \alpha \\ x+\alpha &=150^{\circ}\quad(\because\text { exterior angle of a triangle }) \\ 2 x &=150^{\circ} \\ x &=\dfrac{150^{\circ}}{2}=75^{\circ} \\ y &=\beta \\ y+\beta+\theta &=180^{\circ} \\ 2 y+\theta &=180^{\circ} \\ \text{ But } x+\theta &=180^{\circ}\quad(\because\text { Theorem } 2) \\ 2 y+\theta &=x+\theta \\ 2 y &=x \\ y &=\dfrac{x}{2}=\dfrac{75^{\circ}}{2}=37.5^{\circ} \end{aligned}$


    (c)


    $\begin{aligned} \alpha &=55^{\circ}(\text { radii }) \\ 240^{\circ} &=2(\alpha+\beta)\quad(\because\text { Theorem } 1) \\ \alpha+\beta &=120^{\circ} \\ \beta &=120^{\circ}-\alpha \\ &=120^{\circ}-55^{\circ}\\ &=65^{\circ} \\ y &=\beta \quad \text { (radii) } \\ \therefore y &=65^{\circ} \\ x + \alpha+\beta &=180^{\circ}\quad (\because \text { opposite angles of acyclic quadrilateral )} \\ x &=180^{\circ}-(\alpha+\beta) \\ &=180^{\circ}-\left(55^{\circ}+65^{\circ}\right) \\ &=60^{\circ} \end{aligned}$


    (d)


    $\begin{aligned} \alpha+30^{\circ}+40^{\circ} &=180^{\circ} \\ \alpha&=180^{\circ}-70^{\circ}=110^{\circ} \\ y+\alpha &=180^{\circ}\quad(\because \text { opposite angles of a cyclic quadrilateral }) \\ y &=180^{\circ}-\alpha \\ &=180^{\circ}-110^{\circ}\\ &=70^{\circ} \\ x+\beta &=180^{\circ}\quad(\because\text { opposite angles of a cyclic quadrilateral }) \\ x &=180^{\circ}-\beta \\ &=180^{\circ}-70^{\circ} \\ &=110^{\circ} \end{aligned}$


    (e)


    $\begin{aligned} \alpha&=50^{\circ}\quad(\because\mathrm{radii}) \\ \alpha+\alpha_{1}+50^{\circ} &=180^{\circ} \\ \alpha_{1} &=180^{\circ}-\left(\alpha+50^{\circ}\right) \\ &=180^{\circ}-\left(50^{\circ}+50^{\circ}\right) \\ &=80^{\circ} \\ \beta&=60^{\circ}\quad(\because\mathrm{radii}) \\ \beta+\beta_{1}+60^{\circ} &=180^{\circ} \\ \beta_{1} &=180^{\circ}-\left(\beta+60^{\circ}\right) \\ &=180^{\circ}-\left(60^{\circ}+60^{\circ}\right) \\ &=60^{\circ} \\ \alpha_{1}+\beta_{1}+\theta &=180^{\circ} \\ \theta &=180^{\circ}-\left(\alpha_{1}+\beta_{1}\right) \\ &=180^{\circ}-\left(80^{\circ}+60^{\circ} \right) \\ &=40^{\circ} \\ x &=\gamma \quad(\because\mathrm{radii}) \\ y &=\phi \quad(\because\mathrm{radii}) \\ x+\gamma+\theta &=180^{\circ} \\ 2 x &=180^{\circ}-\theta \\ x &=\dfrac{180^{\circ}-40^{\circ}}{2}=10^{\circ}\\ \theta&=y+\phi\quad(\because\text{exterior angle of a triangle})\\ \theta&=2 y \\ y&=\dfrac{\theta}{2}=\dfrac{40^{\circ}}{2}=20^{\circ} \end{aligned}$


    (f)


    $\begin{aligned} y&=110^{\circ}\quad(\because\text{ exterior angle of a cyclic quadrilateral)}\\ \operatorname{arc} A C&=\operatorname{arc} B D\\ \therefore\alpha&=\beta \\ A C&=B D \\ \therefore A B& ∥ C D \\ \therefore y+z&=180^{\circ} \\ z &=180^{\circ}-y\\ z &=180^{\circ}-110^{\circ}\\ &=70^{\circ}\\ x+z&=180^{\circ}\quad (\because \text{ opposite angles of a cyclic quadrilateral)}\\ x&=180^{\circ}-z\\ &=180^{\circ}-70^{\circ}\\ &=110^{\circ} \end{aligned}$

  3. ABC is an acute triangle inscribed in ⊙O, and OD is the perpendicular drawn O to BC. Prove that ∠ BOD= ∠ BAC.


  4. $\begin{aligned} \text{ Given }&:\triangle A B C \text{ is inscribed in } \odot O and O D \perp B C.\\ \text{ To prove }&:\angle B O D=\angle B A C.\\ \text{ Proof }&: \text{ Draw } OC.\\ &\text { In } \triangle B O D \text { and } \triangle C O D, \\ &B D=D C \quad(\because O D \perp B C) \\ &O D=O D \quad(\because\text { common side }) \\ &O B=O C \quad (\because\text { radii })\\ &\therefore \triangle B O D \cong \triangle C O D\quad(\because\text{ S S S }) \\ \end{aligned}$
    $\begin{aligned} \therefore \alpha &=\beta\\ \angle B O D &=\dfrac{1}{2} \angle B O C\\ \text {But} \quad \angle B A C &=\dfrac{1}{2} \angle B O C \quad(\because\text{ angles substended by arc B C })\\ \therefore \angle B O D &=\angle B A C \end{aligned}$

  5. In ⊙O, two chords AB and CD intersect in the circle at P. Show that ∠ APD= 1/2 ( ∠ AOD + ∠ BOC).


  6. $\begin{aligned} \text{ Given }&: \text{ In }⊙ O, A B \text{ and } C D \text{ intersect at } P.\\ \text { To prove }&: \angle A P D=\dfrac{1}{2}(\angle A O D+\angle B O C)\\ \text { Proof }&:\alpha=\dfrac{1}{2} \angle A O D \quad(\because\text{ angles substended by arc AD })\\ &\quad\beta=\dfrac{1}{2} \angle B O C \quad(\because\text{ angles substended by arc B C } )\\ &\angle A P D=\alpha+\beta \quad\text{(exterior angle of } \triangle A P C )\\ &\quad\quad\quad=\dfrac{1}{2} \angle A O D+\dfrac{1}{2} \angle B O C \\ &\quad\quad\quad=\dfrac{1}{2}(\angle A O D+\angle B O C) \end{aligned}$

  7. Two circles intersect at M and N. From M, diameters MA, MB are drawn in each circle. If A, B are joined to N, prove that ANB is a straight line.


  8. $\begin{aligned} \text{ Given }:&\text{ Two circles intersect at } M \text{ and } N. M A \text{ and } M B \text{ are diameters }.\\ \text{ To prove }:& ANB \text{ is a straight line }.\\ \text{ Proof }:&\alpha=90^{\circ}\quad(\because \text{ angle substended by a díameter })\\ &\beta=90^{\circ} \quad(\because \text{ angle substended by a diameter })\\ &\alpha+\beta=90^{\circ}+90^{\circ}=180^{\circ}\\ &\therefore A N B\text{ is a straight line } \end{aligned}$

  9. OA and OB are two radii of a circle meeting at right angles. From A, B two parallel chords AX, BY are drawn. Prove that AY ⊥ BX.


  10. $\begin{aligned} \text{ Given } : & \text{ In } ⊙ O, O A ⊥ O B \text{ and } A X ∥ B Y . \\ \text { To prove } :& A Y \perp B X . \\ \text { Proof } :& \theta =\alpha+\beta \quad(\because \text { exterior angle of a triange }) \\ &\quad=\alpha+\gamma \quad(\because \beta= \gamma,\quad A X ∥B Y) \\ &\quad=\dfrac{1}{2} \angle A O B+\dfrac{1}{2} \angle A O B \\ &\quad=\angle A O B \\ &\quad=90^{\circ} \\ &\therefore A Y ⊥B X \end{aligned}$

  11. Two circles intersect at R and S. Two straight lines ARB and SCD are drawn meeting one circle at A, C and the other at B, D. Prove that AC ∥ BD. If AB ∥ CD, show that AB= CD.


    $\begin{aligned} \text{ Given }:& \text{ Two circles intersect at } R \text{ and } S.\\ & A R B \text{ and } C S D \text{ are two straight lines }.\\ \text{ To prove }:& A C \| B D\\ & A B \| C D \Rightarrow A B=C D\\ \text{ Proof }:&\text{ Draw } R S.\\ &\alpha+\theta =180^{\circ}\quad(\because \text { opposite angles of a cyclic quadrilateral }) \\ &\quad\quad\theta =\beta \quad(\because\text { exterior angle of a cyclic quadrilateral }) \\ &\therefore \alpha+\beta =180^{\circ} \\ &\therefore A C \| B D \\ & A B \| C D \text { (given) } \\ &\therefore A B C D \text { is a parallelogram. } \\ &\therefore A B=C D \end{aligned}$

Exercise 9.2

  1. In the following figures, O is the centre of circles. Find the values of x and y.

  2. (a)


    $\begin{aligned} O B&=8 \\ \triangle & O B P \text { is a } 30^{\circ}-60^{\circ} \text { right triangle. } \\ \therefore O P&=4, P B=4 \sqrt{3} \\ O Q&=O P\quad(\because\text { given }) \\ \therefore C D&=A B \quad(\because\text { Theorem } 9 ) \\ \dfrac{1}{2} C D&=\dfrac{1}{2} A B\\ C Q&=P B\\ \text {But}\quad C Q&=Q D,A P =P B\quad(\because\text{ Theorem } 7)\\ \therefore x&=P B=4 \sqrt{3} \\ \end{aligned}$


    (b)


    $\begin{aligned} A P&=P B \\ \therefore O P &\perp A B \quad(\because\text { Theorem } 6 ) \\ O P&=5 \quad(\because\text { given }) \\ \triangle & O A P \text { is a } 30^{3}-60^{\circ} \text { right triangle }. \\ \therefore A P&=5 \sqrt{3}, O A=10 \\ y&=A P=5 \sqrt{3} \\ \text { In }\text { rt } &\triangle O Q D, \\ O D^{2} &=O Q^{2}+Q D^{2} \\ Q D^{2} &=O D^{2}-O Q^{2} \\ x^{2}&=O D^{2}-O Q^{2} \\ &=10^{2}-8^{2} \\ &=36 \\ x &=6 \end{aligned}$


    (c)


    $\begin{aligned} (2 x+1+3) \cdot 3&=(13+x-1) \cdot(x-1) \quad(\because\text { Theorem } 12 ) \\ (2 x+4) \cdot 3&=(12+x)(x-1) \\ 6 x+12&=12 x-12+x^{2}-x \\ x^{2}+5 x-24&=0 \\ (x+8)(x-3)&=0 \\ x+8&=0 \text ({ or }) x-3=0 \\ x&=-8 \text ({ or }) x=3 \\ x&=-8 \text { is impossible }. \\ \therefore x&=3 \end{aligned}$

  3. In ⊙O, chord AB is perpendicular to CD at P, AB=16, CP=4, PD= 10. Find the radius.




  4. $\begin{aligned} \quad A B&=16, C P=4, P D=10 \\ \text { Join } & O D \text {. } \\ \text { Draw } & O E \perp A B, O F \perp C D \text {. } \\ A E&=E B=\dfrac{16}{2}=8 \quad(\because\text { Theorem } 7 ) \\ C F&=F D=\dfrac{10+4}{2}=7 \quad(\because\text { Theorem 7 } ) \\ O E P F& \text{ is a rectangle }.\\ \text { Let } E P&=O F=x \\ \therefore E B&=8-x, A P=8+x \\ A P \cdot P B&=C P \cdot P D \quad(\because\text { Theorem } 11 ) \\ (8+x)(8-x)&=4 \times 10 \\ 64-x^{2}&=40 \\ x^{2}&=64-40=24 \\ O F^{2}&=24 \\ \text { In }\text { rightt } &\triangle O F D,\\ O D^{2} & =O F^{2}+F D^{2} \\ & =24+7^{2} \\ &=24+49 \\ &=73 \\ O D & =\sqrt{73} \end{aligned}$

  5. In the figure O is the centre of the concenteric circles and ON ⊥ AB. If OC=10, ON= 8 and OB= 17, find AC.



  6. $\begin{aligned} &O C=10, O N=8, O B=17 \\ &O N \perp A B \\ &\therefore A N=N B\quad(\text { Theorem } 7) \\ \end{aligned}$
    $\begin{aligned} C N^{2} &=O C^{2}-O N^{2} \\ &=10^{2}-8^{2}=100-64=36 \\ C N &=6 \\ N B^{2} &=O B^{2}-O N^{2} \\ &=17^{2}-8^{2}=289-64=225 \\ N B &=15 \\ \therefore A N &=15 \\ A C &=A N-C N \\ &=15-6=9 \end{aligned}$

  7. Prove Theorem 10: Of any twp chords of a circle, the greater chord is nearer to the centre, and conveersely, the chord nearer to the centre is larger.



  8. $\begin{aligned} \text{ Given } :& \text{ In } ⊙O, O E \perp A B \text{ and } OF \perp C D \text{ with } A B>C D.\\ \text{ To prove } :& OE < O F \\ \text{ Proof } :& \text{ Let } A B=2 a, C D=2 b, O E=x, OF =y \text{ and radius } =r.\\ & r^{2}=a^{2}+x^{2} and r^{2}=b^{2}+y^{2}\\ &\therefore a^{2}+x^{2} =b^{2}+y^{2} \\ & b^{2}-a^{2} =x^{2}-y^{2}\\ &\text{ Since } b < a, b^{2}< a^{2}\\ & b^{2}-a^{2} < 0 \\ & x^{2}-y^{2} < 0 \\ & x^{2} < y^{2} \\ & x < y \\ &\therefore O E < O F \end{aligned}$

    Conversely,
    $\begin{aligned} \text{ Given } :& In ⊙ O, O E \perp A B \text{ and } O F \perp C D \text{ with } O E<O F.\\ \text{ To prove } :& A B > C D \\ \text{ Proof } :& \text{ Let } A B=2 a, C D=2 b, O E=x,\\ & O F=y \text{ and radius } =r.\\ & r^{2}=a^{2}+x^{2} \text{ and } r^{2}=b^{2}+y^{2}\\ & \therefore a^{2}+x^{2}=b^{2}+y^{2}\\ & x^{2}-y^{2}=b^{2}-a^{2}\\ &\text{ Since } x < y, x^{2} < y^{2}\\ & x^{2}-y^{2} < 0 \\ & b^{2}-a^{2} < 0 \\ & b^{2} < a^{2} \\ & b < a \\ &\therefore a >b \\ &\therefore \quad A B >C D \end{aligned}$

  9. Let P be a point inside a circle.AB is the diameter passes through P and CPD is the chord perpendicular to AB. Show that CD is the shortest of all chords passing through P.



  10. $\begin{aligned} \text{ Given } :& \text{ In } ⊙ O, A B \text{ is a diameter }.\\ &C D \perp A B \text{ and }X Y \text{ is any line passing through } P.\\ \text{ To prove } :& C D<X Y\\ \text{ Proof } :& \text{ Draw } O Q \perp X Y.\\ &\text{ In right } \triangle O Q P,\\ & O Q < O P\\ & \therefore X Y\rangle C D \quad(\because\text{ Theorem }10)\\ & A B \text{ is the longest chord }.\\ & \therefore C D<X Y<A B.\\ & \therefore C D \text{ is the shortest of all chords passing through } P. \end{aligned}$

  11. Through a point P in a circle, the longest chord that can be drawn is 10cm long and the shortest chord is 6cm long. What is the radius of the circle and how far is P from the centre?



  12. $\begin{aligned} \text{ Given } :& \text{ In } ⊙ O, A B \text{ is the longest chord and } C D \text{ is the shortest chord }.\\ &\text{ The two chords intersest at } P\text{ and } O P \perp C D. \\ \text{ To find } :& O C, O P \\ \text{ Solution } :&\text{ Since the longest chord is the diameter },\\ & A B=10 \mathrm{~cm} \\ &\text { radius }=O C=5 \mathrm{~cm} \\ & C D=6 \mathrm{~cm} \quad (\because\text { given })\\ &\text{ Draw } C D \perp A B.\\ & C P=P D=\dfrac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm} \\ &\text { Let } O P=x \mathrm{~cm} \\ &\text { AP.PB }=C P \cdot P D\quad(\because\text { Theorem 11 }) \\ &(5-x)(x+5) =3\times3 \\ & 25-x^{2} =9 \\ & x^{2} =16 \\ & x =4 \\ & O P =4 \mathrm{~cm}\\ &\therefore P \text{ is } 4 \mathrm{~cm} \text{ far from the centre }.\\ \end{aligned}$

  13. In ⊙O, chord AB and CD are equal and intersect in the circle at E such that AE< EB and CE < ED. Show that △BDE is isosceles with base BD.



  14. $\begin{aligned} \text{ Given } :& \text{ In } ⊙ O, A B=C D.\\ & A B \text{ and } C D \text{ intersect at E }\\ &\text{ such that } A E < E B \text{ and } C E<E D\\ \text{ To prove } :& \triangle B D E \text{ is isosceles with base } B D\\ \text{ Proof } :& \text{ Join } O E, O P \perp A B \text{ and } O Q \perp C D.\\ & A P=P B \quad(\because\text { Theorem 7 }) \\ & C Q=Q D \quad(\because\text { Theorem 7 }) \\ & B u t \quad A B=C D \\ &\dfrac{1}{2} A B=\dfrac{1}{2} C D\\ & P B=Q D\\ &\text{ In right } \triangle O P E \text{ and } \triangle O Q E,\\ & O P =O Q \quad(A B=C D) \\ & O E =O E \quad(\because\text { common side }) \\ &\angle O P E =\angle O Q E=90^{\circ} \\ &\therefore \triangle O P E \cong \triangle O Q E\quad(\because\text { RHL }) \\ &\therefore E P =E Q \\ & E P+P B =E Q+Q D \\ & E B =E D\\ &\therefore \triangle B D E \text { is isosceles with base } BD. \end{aligned}$

  15. In ⊙O,congruent chords AB and CD are produced to meet at P. Prove that △PAC is isosceles .



  16. $\begin{aligned} \text { Given } :& \text { In } ⊙ O, A B=C D \text {. } \\ \text { To prove } :& \triangle P A C \text { is isosceles. } \\ \text { Proof } :& \text { Join OP }, OE \perp A B, O F \perp C P . \\ & A E=E B \quad(\because\text { Theorem } 7)\\ & C F=F D \quad(\because\text { Theorem } 7) \\ & A B=C D\quad(\because\text {given})\\ &\dfrac{1}{2} A B=\dfrac{1}{2} C D\\ & A E=C F--- \text { (1) } \\ & O E=O F \quad(\because\text { Theorem } 9) \\ &\text { In } \triangle O P E \text { and } \triangle O P F, \\ & O P=O P\quad(\because\text { common side }) \\ & O E=O F \quad(\because\text { proved }) \\ &\angle O E P=\angle O F P=90^{\circ} \\ &\triangle O P E \cong \triangle O P F\quad(\because\text { RHL } ) \\ &\therefore P E=P F --- \text { (2) } \\ &\text { Eqn: (1) }+(2), \\ & P E+A E=P F+C F \\ &\therefore P A=P C \\ &\therefore \triangle P A C \text { is isosceles. } \end{aligned}$

  17. In ⊙O,AB and BC are equal chords.OV ⊥ AB, and OU ⊥ BC. Prove that B is the midpoint of arc VU.




  18. $\begin{aligned} \text { Given } :& \text { In } ⊙ O, \quad A B=B C \text {, } \\ & O V \perp A B \text {, } O U \perp B C \text {. } \\ \text { To prove } :& B \text { is the midpoint of } \operatorname{arc} V U \text {. } \\ \text { Proof } :& A B=B C \\ &\dfrac{1}{2} A B=\dfrac{1}{2} B C \\ & B D=B E \\ &\text { Join } O B \text {. } \\ &\text { In } \triangle B D O \text { and } \triangle B E O \text {, } \\ & B D=B E\quad(\because \text { proved ) } \\ & O B=O B \quad\text { ( } \because \text { common side) } \\ &\angle B D O=\angle B E O=90^{\circ} \\ &\triangle B D O \cong \triangle B E O \quad\text { (RHL) } \\ &\therefore \alpha=\beta \\ &\text { arc VB }=\operatorname{arc BU} \\ &\therefore B \text { is the midpoint of } \text { arc VU.} \end{aligned}$

  19. In parallelogram PQRS, PQ=5cm, PR=8cm, QS=6cm. Calculate the lengths of AR and BR.



  20. $\begin{aligned} \text{ Given } :& In parallelogram P Q R S,\\ & S R=P Q=5 \mathrm{~cm}, P R=8 \mathrm{~cm}, O S=6 \mathrm{~cm}\\ \text{ To find } :& A R, B R\\ \text{ Solution } :& P C=C R=\dfrac{8 \mathrm{~cm}}{2}=4 \mathrm{~cm} \\ & S C=C Q=\dfrac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm} \\ & P C \cdot C A=S C \cdot C Q \quad(\because\text { Theorem 11 }) \\ & 4 \cdot(4-A R)=3 \times 3 \\ & 16-4 A R=9 \\ & 4 A R=7 \\ & A R=\dfrac{7}{4}=1.75 \mathrm{~cm} \\ & S R \cdot B R=P R \cdot A R \quad(\because\text { Theorem 12 })\\ & 5 \cdot B R=8 \times \dfrac{7}{4} \\ & B R=\dfrac{14}{5}=2.8 \mathrm{~cm} \end{aligned}$

  21. Chords AB and CD intersect at E and AE=EB. A semicircle is drawn with diameter CD. EF, perpendicular to CD, meets this semicircle at F. Prove that AE=EF.



  22. $\begin{aligned} \text { Given } :& \text { Chords } A B \text { and } C D \text { intersect at } E \text { and } A E=E B \\ & C F D \text { is a semicircle and } E F \perp C D \\ \text { To prove } :& A E=E F \\ \text { Proof } :&\text { Join } C F \text {, FD. } \\ & \angle C F D=90^{\circ} \quad(\because\text { angle substended by a diameter }) \\ & E F \perp C D \text { and } \angle C F D=90^{\circ}, \\ & E F^{2}=C E \cdot E D \quad(\because\text { Theorem 6 }) \\ & A E \cdot E B=C E \cdot E D \quad(\because\text { Theorem } 11) \\ &\therefore A E^{2}=C E \cdot E D \quad(\because A E=E B) \\ &\therefore A E^{2}=E F^{2} \\ &\therefore A E=E F \end{aligned}$

  23. ⊙O and ⊙P intersect at A and B. Show that OP is the perpendicular bisector of common chord AB.



  24. $\begin{aligned} \text{ Given } :& ⊙ O \text{ and } ⊙ P \text{ intersect at } A and B.\\ \text{ To prove } :& O P \perp A B \text{ and } A D=B D.\\ \text{ Proof } :& \text{ Join } O A, O B, P A, P B.\\ &\text { In } \triangle O A P \text { and } \triangle O B P, \\ & O A=O B \quad\text { (radii) } \\ & P A=P B \quad\text { (radii) } \\ & O P=O P \quad\text { (common side) } \\ &\therefore \triangle O A P \cong \triangle O B P\quad({S S S}) \\ &\therefore \alpha_{1}=\alpha_{2}, \\ &\text { In } \triangle O A D \text { and } \triangle O B D, \\ & O A=O B \quad\text { (proved) } \\ & O D = O D\quad(\text { common side }) \\ &\alpha_{1}=\alpha_{2} \quad(\text { proved }) \\ &\therefore \triangle O A D \cong \triangle O B D\quad\text({S A S}) \\ &\therefore \angle O D A=\angle O D B \\ &\angle O D A+\angle O D B=180^{\circ} \\ &\therefore \angle O D A=\angle O D B=\dfrac{180^{\circ}}{2}=90^{\circ} \\ &\therefore O P \perp A B \text { and } A D=B D . \end{aligned}$