Notes-Class-9-Mathematics-2-Chapter-6-Circle-Maharashtra Board

Circle :

The set of points in a plane which are equidistant from a fixed point in the plane is called a circle.

Terms related with a circle :

• The fixed point is called the centre of the circle. (In fig. Point P)
• The segment joining the centre of the circle and a point on the circle is called a radius of the circle. (In fig. seg PA)
• The distance of a point on the circle from the centre of the circle is also called the radius of the circle.
• The segment joining any two points of the circle is called a chord of the circle. (In fig. seg BC)
• A chord passing through the centre of a circle is called a diameter of the circle. (In fig. seg CD)
• A diameter is a largest chord of the circle.

Properties of the chord of the circle :

Theorem : The perpendicular drawn from the centre of a circle on its chord bisects the chord.

Given : seg AB is a chord of a circle with centre O.

seg OP ⊥ chord AB

To prove : seg AP ≅ seg BP

Proof : Draw seg OA and seg OB

In Δ OPA and Δ OPB

∠ OPA ≅ ∠ OPB                             . . . . . . . . . . . seg OP ⊥ chord AB

seg OP ≅ seg OP                  . . . . . . . . . . . . common side

hypotenuse OA ≅ hypotenuse OB   . . . . . . . . . . . radii of the same circle

∴ Δ OPA ≅ Δ OPB                          . . . . . . . . . hypotenuse side theorem

seg PA ≅ seg PB                            . . . . . . . . . . . . c.s.c.t.

Theorem : The segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord.

Given : seg AB is a chord of a circle with centre O and P is the midpoint of chord AB of the circle. That means seg AP ≅ seg PB.

To prove : seg OP ⊥ chord AB

Proof : Draw seg OA and seg OB.

In Δ AOP and Δ BOP

seg OA ≅ seg OB         …….radii of the same circle

seg OP ≅ seg OP          …….common sides

seg AP ≅ seg BP          …….given

∴ Δ AOP ≅ Δ BOP         …….SSS test

∴ ∠ OPA ≅ ∠ OPB         …….c.a.c.t. . . .(I)

Now ∠ OPA + ∠ OPB = 180° …….angles in linear pair

∴ ∠ OPB + ∠ OPB = 180°      …….from (I)

∴ 2∠ OPB = 180°

∴ ∠ OPB = 90°

∴ seg OP ⊥ chord AB

Properties of congruent chords :

Relation between congruent chords of a circle and their distances from the centre :

Theorem : Congruent chords of a circle are equidistant from the centre of the circle.

Given : In a circle with centre O

chord AB ≅ chord CD

OP ⊥ AB, OQ ⊥ CD

To prove : OP = OQ

Construction : Join seg OA and seg OD.

Proof :

AP =  AB, DQ =  CD     …...perpendicular drawn from the centre of a circle to its chord bisects the chord.

AB = CD             …….. given

∴ AP = DQ

∴ seg AP ≅ seg DQ       ……. (I)…….segments of equal lengths

In right angled Δ APO and right angled Δ DQO

seg AP ≅ seg DQ          ……….from (I)

hypotenuse OA ≅ hypotenuse OD            ……..radii of the same circle

∴ Δ APO ≅ Δ DQO        ……..hypotenuse side theorem

seg OP ≅ seg OQ         ……..c.s.c.t.

∴ OP = OQ          …………..Length of congruent segments.

Congruent chords in a circle are equidistant from the centre of the circle.

Theorem : The chords of a circle equidistant from the centre of a circle are congruent.

Given : In a circle with centre O

seg OP ⊥ chord AB

seg OQ ⊥ chord CD 

and OP = OQ

To prove : chord AB ≅ chord CD

Construction : Draw seg OA and seg OD.

Proof :

In right angled Δ OPA and right Δ OQD

hypotenuse OA ≅ hypotenuse OD            ………(Radii of the same circle)

seg OP ≅ seg OQ                  …….given

∴ Δ OPA ≅ Δ OQD                ……….(Hypotenuse side test of congruence)

∴ seg AP ≅ seg QD               ………c.s.c.t.

∴ AP = QD          ……….(I)

But AP =  AB, and DQ =  CD       …….perpendicular drawn from the centre of a

circle to its chord bisects the chord.

and AP = QD                ……..from (I)

∴ AB = CD

∴ seg AB ≅ seg CD

Note that both the theorems are converses of each other

Remember : Congruent chords of a circle are equidistant from the centre of the circle. The chords equidistant from the centre of a circle are congruent.

 Incircle of a triangle :

The circle which touches all the sides of a triangle is called incircle of the triangle and the centre of the circle is called the incentre of the triangle.

In fig., bisectors of all angles of a Δ ABC meet in the point I. Perpendiculars on three sides are drawn from the point of concurrence.

IP ⊥ AB, IQ ⊥ BC, IR ⊥ AC

We know that, every point on the angle bisector is equidistant from the sides of the angle.

Point I is on the bisector of ∠ B. ∴ IP = IQ.

Point I is on the bisector of ∠ C ∴ IQ = IR

∴ IP = IQ= IR

That is point I is equidistant from all the sides of Δ ABC.

∴ if we draw a circle with centre I and radius IP, it will touch the sides AB, AC,

BC of Δ ABC internally.

This circle is called the Incircle of the triangle ABC.

Circumcircle of the triangle :

The circle passing through all the vertices of a triangle is called circumcircle of the triangle and the centre of the circle is called circumcentre of the triangle.

In the figure, the perpendicular bisectors of two sides of Δ PQR intersect at point C. So C is the point of concurrence of perpendicular bisectors. Point C is the circumcentre of Δ PQR.

Circumcircle :

Point C is on the perpendicular bisectors of the sides of triangle PQR. Join PC, QC and RC.

We know that, every point on the perpendicular bisector is equidistant from the end points of the segment.

Point C is on the perpendicular bisector of seg PQ.

∴ PC = QC  ... (1)

Point C is on the perpendicular bisector of seg QR.

∴ QC= RC  ... (2)

∴PC = QC = RC ... [From (1) and (2)]

∴ the circle with centre C and radius PC will pass through all the vertices of Δ PQR. This circle is called as the circumcircle.

To draw the circumcircle of a triangle :

Construct Δ DEF such that DE = 4.2 cm, ∠ D = 60°, ∠ E = 70° and draw circumcircle of it. Draw rough figure. Write the given measures.

Steps of construction :

• Draw Δ DEF of given measures.
• Draw perpendicular bisectors of any two sides of the triangle.
• Name the point of intersection of perpendicular bisectors as C.
• Join seg CF.
• Draw circle with centre C and radius CF.

Type of triangle	Position of incenter	Position of circumcenter
1-Equilateral triangle	Inside the triangle	Inside the triangle
2-Isosceles triangles	Inside the triangle	Inside, outside, on the triangle
3-Scalene triangle	Inside the triangle	Inside, outside or on the triangle
4-Acute angled triangle	Inside the triangle	Inside the triangle
5-Right angled triangle	Inside the triangle	Midpoint of hypotenuse
6-Obtuse angled triangle	Inside the triangle	Outside the triangle

Remember :

• Incircle of a triangle touches all sides of the triangle from inside.
• For construction of incircle of a triangle we have to draw bisectors of any two angles of the triangle.
• Circumcircle of a triangle passes through all the vertices of a triangle.
• For construction of a circumcircle of a triangle we have to draw perpendicular bisectors of any two sides of the triangle.
• The perpendicular bisectors and angle bisectors of an equilateral triangle are coincident.
• The incentre and the circumcentre of an equilateral triangle are coincident.
• Ratio of radius of circumcircle to the radius of incircle of an equilateral triangle is 2 : 1