Theorems Related To Circle

Radius is the line segment that joins center and circumference of a circle.

So, OA, OC, OP, OQ and OB are the radii.

(i) Infinite

The circle is a collection of infinite number of points lying on a plane, each of the points is equidistant from a fixed point on that plane.

(ii) Diameter

A diameter is the greatest chord of any circle.

(iiii) Segment

Any chord of a circle divide the circle in two parts, the one with major area is known as major segment and the other part is known as minor segment.

(iv) Origin

As you can see in this image, Diameter has to pass through origin.

(v) equal

Equal arcs subtend equal segments and sectors.

(vi) radii

A sector is the region enclosed by an arc and two radii as shown below.

OAB is a sector.

(vii) Greater than radius

(i) True

Plane Figure, A figure drawn on a 2-D plane.

(ii) True

The segment of a circle is the region bounded by a chord and the arc subtended by the chord.

As you can see in the figure the segment is a plane figure.

(iii) True

A sector is the part of a circle enclosed by two radii of a circle and their intercepted arc.

As you can see in the figure the Sector is a plane figure.

(iv) True

A line segment is a distance between two points.

A chord is a distance between two points on the circumference.

(v) False

Arc is a segment of differential Curve.

(vi) False

There are infinite number of Chords of same length.

(vii) False

Concentric circles are circles with a common center. As you can see in this image We can draw infinite circles.

(viii) True

Two circles are congruent if they have the same size. The size can be measured as the radius, diameter or circumference. They can overlap

Given radius = 5cm, AB = 5cm

Perpendicular from the center of the circle to any Chord bisects it in two line segments

So, AD = BD

In ODB

OB = 5cm

BD = 4cm

Using Pythagoras Theorem

base² + Height² = Hypotenuse²

⇒ OD² + DB² = OB²

⇒ OD² = 25-16

⇒ OD² = 9

⇒ OD = 3cm

Given Length of the diameter = 26cm, Distance of Chord PQ From center = 5cm

Radius

Radius = 13cm

In O PD , Using Pythagoras Theorem

Hypotenuse² = base² + Height²

⇒ OP² = OD² + PD²

⇒ 13² = 5² + PD²

⇒ PD² = 169-25

⇒ PD² = 144

⇒ PD = 12cm

⇒ PQ = 2 PD

⇒ PQ = 24cm

Given PQ = 4cm, Distance from the Center is 2,1cm

PQ = 2 PD

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ PD

⇒ PD = 2cm

In O PD , Using Pythagoras Theorem

⇒ OD² + PD ² = OP²

⇒ OP² = 2.1² + 2²

⇒ OP² = 8.41

⇒ OP = 2.9cm

Let Smaller Chord is EF, the other chord is BC, distance from the centre o to smaller chord is OG.

⇒ BC = 8cm

⇒ CD

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ CD = 4cm

⇒ EF = 6cm

⇒ EG

⇒ EG = 3cm

⇒ OG = 4cm

In OEG, Using Pythagoras Theorem

⇒ OE² = OG² + EG²

⇒ OE² = 16 + 9

⇒ OE² = 25

⇒ OE = 5cm

In OCD,Using Pythagoras Theorem

⇒ OC² = CD² + OD²

⇒ 5² = 4² + OD²

⇒ 25 = 16 + OD²

⇒ OD² = 9

⇒ OD = 3cm

Given AB = 48cm, OP = 7cm

AP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ AP = 24cm

In OAP, Using Pythagoras Theorem

⇒ OA² = AP² + OP²

⇒ OA² = 24² + 7²

⇒ OA² = 576 + 49

⇒ OA² = 625

⇒ OA = 25cm

Given, AB = 6cm, PC = 2cm

AP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ AP

⇒ AP = 3cm

⇒ OA = OC

⇒ OA = OP + CP ………..(1)

In OAP, Using Pythagoras Theorem

OA² = OP² + AP²

⇒ (OP + PC)² = OP² + 9 (from eq.(1))

⇒ OP² + PC² + 2(OP)(PC) = OP² + 9

⇒ PC² + 2(OP)(PC) = 9

⇒ 4 + 2(OP)(2) = 9

⇒ 2(OP)(2) = 5

⇒ OP

Radius = OC = OP + PC

⇒ OC = 2 +

⇒ OC = 2.25cm

Given, Concentric Circles, CD and AB chords.

To prove: AC = DB

Construction: OP is a perpendicular bisector of CD and AB.

In OAP and OBP, We have

AP = BP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

In OCP and ODP

CP = DP

⇒ CA + AP = BP + DB

⇒ CA = DB + BP-AP

⇒ CA = DB + 0

⇒ CA = DB

Given, AB and CD are Diameters.

To prove: OA = OB, OC = OD

Construction: Point D joined with B, Point A joined with C.

DAC = 90⁰ (Rectangle)

ACB = 90⁰ (Rectangle)

CBD = 90⁰ (Rectangle)

BDA = 90⁰ (Rectangle)

ACBD is Rectangle, So AD = CB, AD BCand BD = AC,BD AC

In OCB and ODA

OCB = ODA (Interior angles, BC AD)

BC = AD (Rectangle)

OBC = OAD (Interior angles, BC AD)

BCA SA Congruency

In OCB ODA

Hence Using CPCT, OA = OB, OD = OC

Hence Proved.

Given, Two Circles with Centers X and Y intersect each other at point A and B. S is a mid-point of XY, AS is perpendicular to PQ.

To Prove: AP = AQ

Construction: XM perpendicular to chord PA. And YN perpendicular to chord AQ of respective circles.

Since SA is also perpendicular to PQ (given)

So, all these perpendiculars to the same line are parallel to each other.

SX = SY

AM = AN

So,

AP = AQ

Given, AB = 10cm, CD = 24cm, PQ = 17cm

AP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

AP

AP = 5cm

CQ

⇒ CQ

⇒ CQ = 12cm

In OAP, Using Pythagoras Theorem

OA² = AP² + OP²

⇒ OA² = 5² + OP²

⇒ OA² = 25 + OP² ………… (1)

⇒ In OCQ, Using Pythagoras Theorem

⇒ OC² = CQ² + OQ²

⇒ OC² = 12² + OQ²

⇒ OC² = 144 + OQ² ………. (2)

⇒ OC = OA

⇒ 144 + OQ² = 25 + OP²

⇒ OP²-OQ² = 119

⇒ (OP-OQ)(OP + OQ) = 119 (using (A² + B² = (A + B)(A-B))

⇒ OQ + OP = 17 ………. (3)

⇒ (OP-OQ)17 = 119

⇒ OP-OQ = 119/17

⇒ OP-OQ = 7 …….. (4)

Eq.3 + Eq.4

⇒ 2OP = 24

⇒ OP = 12cm

⇒ OP-OQ = 7

⇒ OQ = OP-7

⇒ OQ = 12-7

⇒ OQ = 5cm

In OAP, Using Pythagoras Theorem

⇒ OC² = 144 + OQ²

⇒ OC² = 144 + 25

⇒ OC = 13cm

Given, CD is parallel to PQ, Two Circles with centers P and Q intersect at point A and B.

To Prove: 2PQ = CD

Construction: PE and QE are the Perpendiculars on the Chord CD From centers P and Q respectively.

⇒ DE = EA

⇒ CF = AF

⇒ DE + EA + AF + FC = DC

⇒ 2EA + 2AF = DC

⇒ 2(EA + AF) = DC ……..(1)

PQ and CD are parallel lines and PE and QE are the perpendiculars.

So, AEP = 90⁰,AFQ = 90⁰,EPQ = 90⁰,AFQ = 90⁰

EPQF is a rectangle

So, EF = PQ ……… (2)

⇒ 2(EA + AF) = DC

⇒ 2EF = DC

⇒ 2PQ = DC

Given, AC = AB, BAF = CAF

To prove: FB = FC

In ABF and CAF

AC = AB (Given)

BAF = CAF (Given)

AF = AF (Common)

ABF CAF

BFA = CFA (CPCT)

FB = FC (CPCT)

AE is a perpendicular bisector of chord BC, So, It has to pass through the center of the circle.

Given, OF is angle bisector of AFC.

Construction: OQ ⊥ AB and OP ⊥ CD

In ΔOFQ and ΔOFP
∠OFQ = ∠OFP (given)
OF = OF(Common)
∠OQF = ∠OPF(Construction)

AAS Congruency.
ΔOPR ≅ΔOPQ.

∴ OR = OQ (C.P.C.T)

Hence AB = CD

To Prove: DH>EG

Construction: B is the center of the circle, CD and EF are the chords, BH and BG are the perpendicular bisector of CD and EF respectively.

Given BH<BG As it given, CD is nearer to center than EF.

In BDH

⇒ BD² = BH² + DH² …………… (1)

In BEG

⇒ BE² = BG² + EG² ………… (2)

⇒ BD = BE

From Eq1 and Eq2

⇒ BH² + DH² = BG² + EG²

⇒ BH<BG

⇒ BH²<BG²

⇒ BH²-BG²<0

From Eq.3

⇒ BH² + DH² = BG² + EG²

⇒ BH²-BG² = EG²-DH²

⇒ EG²-DH²<0

⇒ EG²<DH²

⇒ EG<DH

To Prove CD<BG

Let, Center is O, Two chords are BG and CD, OM and OE are perpendicular bisector of BG and CD respectively.

In OME,

⇒ OE>OM

In OMB, Using Pythagoras Theorem

⇒ OB² = OM² + BM² ………. (1)

In OCE

⇒ OC² = OE² + CE² …………….. (2)

⇒ OC = OB

⇒ OM² + BM² = OE² + CE²

⇒ OE>OM

So, BM>CE

⇒ BG>CD

As Chord goes near, its length increases.

In AOB and COD

AB = CD

OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

COD = 60⁰

Given radius = 13cm, PQ = 10cm

⇒ PD

⇒ PD

⇒ PD = 5cm

Using Pythagrus theorem in O PD

⇒ OP² = OD² + PD ²

⇒ 13² = OD² + 5²

⇒ OD² = 169-25

⇒ OD² = 144

⇒ OD = 12cm

In AOB and COD

AB = CD

OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

OAB = OCD = OBA = ODC

⇒ EA

⇒ FC

⇒ AB = CD

So, EA = FC

In AOE and COE

⇒ EA = FC

⇒ EAO = FCO

⇒ OA = OC

SAS Congruency.

So, By CPCT OE = OF = 4cm

Let Parallel Chords are AB and CD. OE and OF are perpendicular bisector of AB and CD respectively.

Given, OB = OA = OD = OC = 10cm, AB = CD = 16cm

⇒ DF

⇒ DF

⇒ DF = 8cm

⇒ EB

⇒ EB

⇒ EB = 8cm

In AOE

⇒ OA² = AE² + OE²

⇒ OE² = OA²-AE²

⇒ OE² = 10²-8²

⇒ OE² = 100-64

⇒ OE² = 36

⇒ OE = 6cm

As we can see from the previous question both the triangles are congruent.

So, OE = OF

Hence Distance between parallel lines = OE + OF = 6 + 6 = 12cm

In OAP and OBP

⇒ AP = BP

⇒ OA = OB

⇒ OP = OP

In OCP and ODP

⇒ CP = DP

⇒ CA + AP = BP + DB

⇒ CA = DB + BP-AP

⇒ CA = DB + 0

⇒ CA = DB

So, BD = 5cm

(i) False

There is no Circle passes through the 3 Collinear points.

(ii) True

Only one Circle Passes through three Collinear Points.

(iii) False

You can see in this image, AB and AC are Situated on the Opposite side of the radius OA, Angles OAB and OCA are not Equal. It holds only for Equal chords.

(i) 1:1

In AOB and COD

AB = CD

OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

OAB = OCD = OBA = ODC

(ii) Passes through the origin of that Circle.

As you can see in this image The perpendicular bisector passes through the centre.

Let Center of the Circles are A and B. CD is a common Chord of the circle. AB is the perpendicular bisector of the chord CD.

If AB is a perpendicular bisector of CD then it should passes through both the centers.

So, AB is the distance that we need to calculate.

Given, AC = 10cm, CD = 12cm

⇒ CM

⇒ CM

⇒ CM = 6cm

In ACM

⇒ AC² = AM² + CM²

⇒ AM² = AC²-CM²

⇒ AM² = 100-36

⇒ AM² = 64

⇒ AM = 8cm

In BCM

⇒ BC² = BM² + CM²

⇒ BM² = BC²-CM²

⇒ BM² = 100-36

BM² = 64

BM = 8cm

AB = 16cm

Given, AC = AB, BAF = CAF, AB and AC are two equal Chords of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.

In ABF and CAF

⇒ AC = AB

⇒ BAF = CAF

⇒ AF = AF

⇒ ABF CAF

⇒ BFA = CFA

⇒ FB = FC

AE is a perpendicular bisector of chord BC.

In ΔABF, by Pythagoras theorem,

⇒ AB² = AF² + BF²

⇒ BF² = 6² - AF² .............(1)

In OBF

⇒ OB² = OF² + BF²

⇒ 5² = (5 - AF)² + BF²

⇒ BF² = 25 - (5 - AP)² ...........(2)

Equating (1) and (2), we get

⇒ 6² - AF² = 25 - (5 - AF)²

⇒ 11 - AF² = -25 - AF² + 10AF

⇒ 36 = 10AF

⇒ AF = 3.6 cm

Putting AF in (1), we get

⇒ BF² = 6² - (3.6)² = 23.04

⇒ BF = 4.8 cm

⇒ BC = 2BF = 2 × 4.8 = 9.6 cm

In AOB and COD

⇒ AB = CD

⇒ OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

AB = 6cm

⇒ COD = 60⁰

⇒ AE

⇒ AE

⇒ AE = 3cm

In AOE and BOE

⇒ OA = OB

⇒ OE = OE

⇒ AE = BE

Hence Using SSS congruency

⇒ AOE BOE

⇒ AOE = BOE

⇒ AOE = 30⁰

⇒ Sin30

⇒

⇒ OA = 2AE

⇒ OA = 2(3)

⇒ OA = 6cm

Given: OP = 3cm, Radius = 5cm

Let OA, OC radius of the circle, OM is perpendicular Bisector Passes through the centre O of Chord CD. OP is a Perpendicular Bisector of Chord AB passes through the Centre O.

In OPM

OMP = 90⁰

Using, Pythagoras Theorem

⇒ OP² = OM² + PM²

⇒ OP>OM

We have proved in Question 14, That Nearer chord is greater than the other.

It shows us that

AB<CD

Hence The Least chord passes through point P is AB, OP is Perpendicular bisector of the chord.

In OPA

Using, Pythagoras Theorem

⇒ OA² = OP² + AP²

⇒ 5² = 3² + AP²

⇒ AP² = 25-9

⇒ AP² = 16

⇒ AP = 4cm

Using above Theorem

AB = 8cm

Given: PQ = 5cm, CD is parallel to PQ, Two Circles with centers P and Q intersect at point A and B.

Construction: EF and FQ are Perpendicular bisector drawn from P and Q Respectively.

The Perpendicular from The Centre to the chord, bisects the chord.

⇒ DE = EA

⇒ CF = AF

⇒ DE + EA + AF + FC = DC

⇒ 2EA + 2AF = DC

⇒ 2(EA + AF) = DC ………..(1)

EFPQ (Given)

PEA = 90^o (Construction)

QFA = 90^o (Construction)

Using Interior Angle Theorem,

PEA + BPE = 180

BPE = 90^o

PQFE is a rectangle.

So, EF = PQ …………… (2)

⇒ 2(EA + AF) = DC

⇒ 2EF = DC

⇒ 2PQ = CD

CD = 10cm