# Summarizing the discussion regarding circle

1.â€‹

*Property:*In a circle (or congruent circles) equal chords are made by equal arcs. {OP = OQ} = {Oâ€™R = Oâ€™S) PQ = RS and PQ = RS

â€‹2.

â€‹

Property: Equal arcs (or chords) subtend equal angles at the centre PQ = AB (or PQ = AB) âˆ POQ =âˆ AOB

Property: Equal arcs (or chords) subtend equal angles at the centre PQ = AB (or PQ = AB) âˆ POQ =âˆ AOB

3.

4.

Property: The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord. AD = DBOD âŠ¥ AB

5.

6.
âˆ´ AB = PQ
âˆ´ OD = OR

7.
âˆ´ OD = OR
âˆ´ AB = PQ

8.

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12.

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14.

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24.

Property: The perpendicular from the centre of a circle to a chord bisects the chord i.e., if OD âŠ¥ AB (OD is perpendicular to AB).

4.

Property: The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord. AD = DBOD âŠ¥ AB

5.

*Property:*Perpendicular bisector of a chord passes through the centre i.e., OD âŠ¥ AB and AD = DB âˆ´ O is the centre of the circle

6.

*Property:*Equal chords of a circle (or of congruent circles) are equidistant from the centre

7.

*Property:*Chords of a circle (or of congruent circles) are equidistant from the centre

8.

*Property:*The angle subtended by an arc (the degree measure of the arc) at the centre of a circle is twice the angle subtended by the arc at any point on the remaining part of the circle. mâˆ AOB = 2m âˆ ACB.

9.

*Property:*Angle in a semicircle is a right angle.

10.

*Property:*Angles in the same segment of a circle are equal i.e., âˆ ACB = âˆ ADB

11.

*Property:*If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, then the four points lie on the same circle. âˆ ACB = âˆ ADB

âˆ´ Points A, C, D, B are concyclic i.e., lie on the circle

12.

*Property:*The sum of pair of opposite angles of a cyclic quadrilateral is 180Â°.

âˆ DAB + âˆ BCD = 180Â°

and âˆ ABC + âˆ CDA = 180Â° (Inverse of this theorem is also true)

13.

*Property:*Equal chords (or equal arcs) of a circle (or congruent circles) subtended equal angles at the centre.

AB = CD (or AB = CD) âˆ AOB = âˆ COD

(Inverse of this theorem is also true)

14.

*Property:*If a side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. m

âˆ CDE = m âˆ ABC

15.

*Property:*A tangent at any point of a circle is perpendicular to the radius through the point of contact.

(Inverse of this theorem is also true)

16.

*Property:*The lengths of two tangents drawn from an external point to a circle are equal i.e., AP = BP

17.

*Property:*If two chords AB and CD of a circle, intersect inside a circle (outside the circle when produced at a point E), then AE Ã— BE = CE Ã— DE

18.

*Property:*If PB be a secant which intersects the circle at A and B and PT be a tangent at T then PA Ã— PB = (PT)

^{2}

19.

*Property:*From an external point from which the tangents are drawn to the circle with centre O, then (a) they subtend equal angles at the centre (b) they are equally inclined to the line segment joining the centre of that point âˆ AOP = âˆ BOP and âˆ APO = âˆ BPO

20.

*Property:*If P is an external point from which the tangents to the circle with centre O touch it at A and B then OP is the perpendicular bisector of AB.

OP âŠ¥ AB and AC = BC

21.

Property: If from the point of contact of a tangent, a chord is drawn then the angles which the chord makes with the tangent line are equal respectively to the angles formed in the corresponding alternate segments. In the adjoining diagram.

âˆ BAT = âˆ BCA and âˆ BAP = âˆ BDA

22.

*Property:*The point of contact of two tangents lies on the straight line joining the two centres.

(a) When two circles touch externally then the distance between their centres is equal to sum of their radii, i.e., AB = AC + BC

(b) When two circles touch internally the distance between their centres is equal to the difference between their radii. i.e., AB = AC â€“ BC

23.

*Property:*For the two circles with centre X and Y and radii r

_{1}and r

_{2}. AB and CD are two Direct Common Tangents (DCT), then the length of DCT

24.

Property: For the two circles with centre X and Y and radii r

_{1}and r_{2}PQ and RS are two transverse common tangent, then length of TCT# Some important points

- 1. If three circles, each of radius r, are so kept that each circle touches the other two, then the area of the shaded region is r
^{2}[âˆš3 â€“ (Ï€/2)] which is approximately equal to (4/25) r^{2}. - If four circles, each of the radius r, are so kept that each circle touches the other two, then the area of the shaded region is rÂ² (4 â€“ Ï€) which is equal to (6/7)r
^{2}. - In the diagram given below, if the radius (OA) of the outer circle is R and the radius (OB) of the inner circle is r, then the width (w) of the ring is (R âˆ’ r) and the area of the shaded region is Ï€(R
^{2 }âˆ’ r^{2}) or Ï€(R âˆ’ r) (R + r) or Ï€(w) (R + r)

Example-1

Two identical circles intersect so that their centers, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles, is

Solution

Shaded area = 2 Ã— (area of sector ADC â€“ area of âˆ†ADC)

Hence, option (b)

Example-2

Four points A, B, C and D lie on a straight line in the X â€“ Y plane, such that AB = BC = CD, and the length of AB is 1 cm. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance the ant must traverse to reach the sugar particle (in m) is

- 1 +
- 5

Solution

In the drawn figure, ant will go along APQD since it cannot be within a distance of 1 cm from the repellents kept at B and C.

Also AP = QD

Example-3

Two circles C_{1} and C_{2}, having the same radius of 2 cm and centers at P and Q respectively, interÂsect each other such that the line of centers PQ intersects C_{1} and C_{2} at F and E respectively. EF = 1 cm. The whole assembly is enclosed in a rectangle of the minimum area. The perimeter of the rectangle is

Hence, breadth of the rectangle = 4 cm

And length = 7 cm

Perimeter = 2 Ã— (7 + 4) = 22 cm

Example-4

Semi-circle C

_{1}is drawn with a line segment PQ as its diameter with centre at R. Semicircles C_{2}and C_{3 }are drawn with PR and QR as diameters respectively, both C_{2}and C_{3}lying inside C_{1}. A full circle C_{4}is drawn in such a way that it is tangent to all the three semicircles C_{1}, C_{2}and C_{3}. C_{4}lies inside C_{1}and outside both C_{2}and C_{3}. The radius of C_{4}isSolution

_{4}= r and PQ = k.

â‡’ RS = (k/2) â€“ r

RT = k/4

ST = (k/4) + r

â‡’ r = k/6 = PQ/6

Example-5

On a semicircle with diameter AD, chord BC is parallel to AD. Further each of the chords AB and CD has length 2 units, while AD = 8 units. What is the length of BC?

- 7.5
- 7
- 7.75
- 8

Solution

Example-6

The adjacent sides AB, BC of a square of side â€˜aâ€™ units are tangent to a circle. The vertex D of the square lies on the circumference of the circle. The radius of the circle could be:

- a(2 â€“ âˆš2)
- a(âˆš2 â€“ 1)
- a(âˆš2 + 1.5)
- a(âˆš2 + 1)

Solution

OD = r, OB = râˆš2
Hence, r + râˆš2 = âˆš2 a
Hence, r = a(2 âˆ’âˆš2)

Example-7

M, N, O and P are centres of four interÂsecting circles each having a radius of 15 cm. If AB = 7 cm, CD = 5 cm, EF = 6 cm, GH = 8 cm, what is the perimeter of the quadrilateral MNOP?

- 84 cm
- 45 cm
- 94 cm
- 124 cm

Solution

MN = 30 â€“ 6 = 24 cm
NO = 30 â€“ 5 = 25 cm
OP = 30 â€“ 7 = 23 cm
PM = 30 â€“ 8 = 22 cm
So, the perimeter = 24 + 25 + 23 + 22 = 94 cm

Example-8

AB = BC = AC = CD. Find âˆ a.

- 30Â°
- 60Â°
- 15Â°
- None of these

Solution

âˆ ACB = 60Â° âˆ ACD = 180 â€“ 60 = 120Â° âˆ CAD = âˆ CDA = 30Â° âˆ Î± = 30Â°

Example-9

In the figure shown here QS = SR, QU = SU, PW = WS and ST||RV. What is the value of the

Solution

â‡’ PY = YX

PV = TV (in âˆ†PTS; VW||ST) and QA = AT (in âˆ†QTS:

AU||ST) and QT = VT (in âˆ†VQR; ST||VR)

*VT*)+ (

*VP*)

â‡’

*UX*(

*XY*)

so, in âˆ†PUS,

*UX*

*XY*

*PY*â‡’

â‡’ Area (âˆ†

*SUX*) Area (âˆ†

*PUS*)

â‡’ Area (âˆ†

*SUX*) (âˆ†

*SUX*), then

Area (âˆ†

*PSX*) = Area (âˆ†

*PSU*) â€“ area (âˆ†SUX) (âˆ†

*PQR*)

Example-10

Through T, the mid-point of the side QR of a âˆ†PR, a straight line is drawn to meet PQ produced to S and PR at U, so that PU = PS. If length of UR = 2 units then the length of QS is:

- 2 Units
- Units
- 2 Units
- Cannot be determined

Solution

We have QT = TR and PU = PS and UR = 2 units

Draw RV||PS that meets SU extended at V.

Now, in âˆ†QST and âˆ†TVR

âˆ QST = âˆ TVR (alternate angles as PS||VR) and

âˆ QTS = âˆ VTR

QT = TR

âˆ´ DQST and DTVR are congruent.

âˆ´ QS = VR â€¦ (1)

Now âˆ QST = âˆ PUS = âˆ VUR = âˆ UVR

In âˆ†UVR, âˆ VUR = âˆ RVU

or, RV = UR = 2 â€¦(2)

From (1) and (2)

QS = VR = UR = 2 units

Now, in âˆ†QST and âˆ†TVR

âˆ QST = âˆ TVR (alternate angles as PS||VR) and

âˆ QTS = âˆ VTR

QT = TR

âˆ´ DQST and DTVR are congruent.

âˆ´ QS = VR â€¦ (1)

Now âˆ QST = âˆ PUS = âˆ VUR = âˆ UVR

In âˆ†UVR, âˆ VUR = âˆ RVU

or, RV = UR = 2 â€¦(2)

From (1) and (2)

QS = VR = UR = 2 units

Example-11

ABCD is a parallelogram and P is any point within it. If the area of the Parallelogram ABCD is 20 units, then what is the sum of the areas of the DPAB and DPCD?

- 5 units
- 10 units
- 12 units
- Cannot be determined

Solution

Let AB = CD =
Area of (âˆ†PAB + âˆ†PCD)

Given area of the parallelogram = 20 units

Area of (âˆ†PAB + âˆ†PCD) units

*a*and*x*,*y*be the lengths of the perÂpendiculars from P on AB and CD respectively, then*ax*+*ay**a*(*x + y*)*ah*Given area of the parallelogram = 20 units

Area of (âˆ†PAB + âˆ†PCD) units

Example-12

Nayantara bought a triangular piece of land of area 150 m

^{2}. He took a piece of rope and meaÂsured the two sides of the plot and found the largest side to be 50 m and another side to be 10 m. What is the exact length of the third side?- 32 m

Solution

Now, area of âˆ†ABC = (1/2) (BC) (AD) (Where D is a point on BC such that ADâŠ¥BC)

Now, AD has to be equal to

AD = m = 6 m

âˆ´ BD

^{2}= AB

^{2}â€“ AD

^{2}= 10

^{2}â€“ 6

^{2}= 8

^{2}m

^{2}

â‡’ BD = 8 m and DC = BC â€“ BD = 42 m

^{2}= AD

^{2}+ DC

^{2}= (6)

^{2}+ (42)

^{2}

â‡’ AC =

Example-13

C

_{1}and C_{2}are two concentric circles with radii 5 cm and 9 cm respectively. If A, B and C are points on C_{2}such that AB and AC are tangent to C_{1}at how many points does BC intersect C_{1}?- 0
- 1
- 2
- Cannot be determined

Solution

_{1}and C

_{2}become the incircle and circumcircle of ABC respectively, which will be an equilateral triangle)

Example-14

A rectangle MNOQ is drawn and length â€˜NOâ€™ is extended to point R and a triangle QPR is drawn with QP QM. Angle QRP = 45Â° and side QR S and T are the midpoints of sides QR and PR respectively. If ST = 6 units, the area (in sq. cm) of the rectangle is

- 112
- 144
- 288
- 256

Solution

Since the line joining the mid-points of two sides of a triangle is parallel and equal to half the third side, we have PQ = 2(ST), â‡’ PQ = 12 cm
Since, PQ QM

Now, âˆ ORP = 45Â°

Draw PVâŠ¥ON

In âˆ†RVP, tan 45Â° =

(Since OV = PQ = 12 cm) also, breadth
In âˆ†ROQ, â‡’ Breadth =

â‡’ (

â‡’ 2

â‡’

â‡’

Since
Breadth =

= 16 Ã— 18 = 288 sq. cm.

Now, âˆ ORP = 45Â°

Draw PVâŠ¥ON

In âˆ†RVP, tan 45Â° =

(Since OV = PQ = 12 cm) also, breadth

^{2}+ OR^{2}= (QR)^{2}*x*+ 12 where OR =*x*â‡’ (

*x*+ 12)^{2}= âˆ’*x*^{2}(given that QR =â‡’ 2

*x*^{2}+ 24*x*â€“ 128 = 0â‡’

*x*^{2}+ 12*x*â€“ 24 = 0â‡’

*x*= 4 or âˆ’16Since

*x*> 0,*x*= 4*x*+ 12 i.e., 16 cm, and the area of MNOQ= 16 Ã— 18 = 288 sq. cm.

Example-15

âˆ†ABC has sides AB, AC measuring 2001 and 1002 units respectively. How many such triangles are possible with all integral sides?

- 2001
- 1002
- 2003
- 1004

Solution

Value of BC will lie in between 999 and 3003. Hence 999 < BC < 3003.
So, the total values possible for BC = 2003
Given: OB = OD = a (Semi-major axis)
OA = Ol = b (Semi = minor axis)

**Ellipse**The path of a moving point which moves in such a way that its distance from a fixed point (focus) bears a constant ratio with its distance from a fixed line (directrix)- Area = Ï€ab
- Perimeter = Ï€(a + b)

**Maths behind the formula (Area = Ï€ab)**One way to see why the formula is true is to realize that the ellipse is just a unit circle that has been stretched by a factor â€˜

*a*â€™ in the

*x*-direction and by factor â€˜

*b*â€™ in the

*y*-direction. Hence the area of the ellipse is just

*a*Ã—

*b*times the area of unit circle.