Revision: Geometry >> Circles Maths English Medium Class 9 CBSE

A circle is a closed curve where all points on the boundary (called the circumference) are at the same distance from a fixed point inside it.

• The fixed point inside the circle is called the center (O)

The radius is a straight line segment that connects the center of the circle to any point on its circumference.

Characteristics:

• Symbol: Usually represented as r

• All radii of a circle have the same length

• A circle has infinite radii (one to every point on the circumference)

• The radius is always half the diameter

• Radius = `"Diameter"/"2"`

 The diameter is a straight line segment that passes through the center of the circle and has both endpoints on the circumference.

Characteristics:

• The diameter passes through the center

• A circle has infinite diameters

• The diameter is the longest possible chord of a circle

• The diameter is twice the radius

• Diameter = 2 × Radius and

A chord is a straight line segment that connects any two points on the circumference of the circle.

Characteristics:

• A circle has infinite chords

• The diameter is the longest chord in any circle

• Chords closer to the centre are longer than chords farther from the center

The line segment, joining any two points on the circumference of the circle, is called a chord. 

We have to prove that

AQ = `1/2` (perimeter of ΔABC)

Perimeter of ΔABC = AB + BC + CA

= AB + BP + PC + CA

= AB + BQ + CR + CA

(∵ Length of tangents from an external point to a circle are equal ∴ BP = BQ and PC = CR)

= AQ + AR  ...(∵ AB + BQ = AQ and CR + CA = AR)

= AQ + AQ  ...(∵ Length of tangents from an external point are equal)

= 2AQ

⇒ AQ = `1/2` (Perimeter of ΔABC)

Hence proved.

Given: A circle with centre O and an external point T from which tangents TP and TQ are drawn to touch the circle at P and Q.

To prove: ∠PTQ = 2∠OPQ.

Proof: Let ∠PTQ = xº.

Then, ∠TQP + ∠TPQ + ∠PTQ = 180º   ...[∵ Sum of the ∠s of a triangle is 180º]

⇒ ∠TQP + ∠TPQ = (180º – x)   ...(i)

We know that the lengths of tangent drawn from an external point to a circle are equal.

So, TP = TQ.

Now, TP = TQ

⇒ ∠TQP = ∠TPQ

`= \frac{1}{2}(180^\text{o} - x)`

`= ( 90^\text{o} - \frac{x}{2})`

∴ ∠OPQ = (∠OPT – ∠TPQ)

`= 90^\text{o} - ( 90^\text{o} - \frac{x}{2})`

`= \frac{x}{2} `

`⇒ ∠OPQ = \frac { 1 }{ 2 } ∠PTQ`

⇒ 2∠OPQ = ∠PTQ

Given: TP and TQ are two tangents of a circle with centre O and P and Q are points of contact.

To prove: ∠PTQ = 2∠OPQ

Suppose ∠PTQ = θ.

Now by theorem, “The lengths of a tangents drawn from an external point to a circle are equal”.

So, TPQ is an isoceles triangle.

Therefore, ∠TPQ = ∠TQP

`= 1/2 (180^circ - θ)`

`= 90^circ - θ/2`

Also by theorem “The tangents at any point of a circle is perpendicular to the radius through the point of contact” ∠OPT = 90°.

Therefore, ∠OPQ = ∠OPT – ∠TPQ

`= 90^@ - (90^@ -  1/2theta)`

`= 1/2 theta`

= `1/2` ∠PTQ

Hence, 2∠OPQ = ∠PTQ.

Theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

If in a circle with centre O , arc PQ of a circle subtends ∠POQ at centre and arc PQ subtends ∠PAQ on the remaining part of the circle than  ∠POQ = 2 PAQ.

Theorem: Angles in the same segment of a circle are equal.  
Suppose we join points P and Q and form a chord PQ in the above figures. Then ∠ PAQ is also called the angle formed in the segment PAQP.

∠ POQ = 2 ∠ PCQ = 2 ∠ PAQ 
Therefore, ∠ PCQ = ∠ PAQ.
 Here ∠PAQ is an angle in the segment, which is a semicircle. Also, ∠ PAQ = `1/2`
∠POQ = `1/2 xx = 180° =90° ` 

Theorem:   If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).
In following fig. AB is a line segment, which subtends equal angles at two points C and D. That is ∠ ACB = ∠ ADB 

To show that the points A, B, C and D lie on a circle. let us draw a circle through the points A, C and B. Suppose it does not pass through the point D. Then it will intersect AD (or extended AD) at a point, say E .
If points A, C, E and B lie on a circle, 
∠ ACB = ∠ AEB 
But it is given that ∠ ACB = ∠ ADB. 
Therefore, ∠ AEB = ∠ ADB. 
This is not possible unless E coincides with D. 
Similarly, E′ should also coincide with D.

Given:

O is the circumcenter of triangle ABC.

D is the foot of the perpendicular from O to BC.

So OD ⟂ BC.

We have to prove that   ∠BOD = ∠A

OB = OC = OA

Since O is the circumcenter, all these are radii of the circumcircle of triangle ABC.

OD ⟂ BC ⇒ D is midpoint of BC

In a circle, the perpendicular drawn from the center to any chord bisects that chord.

BC is a chord and OD ⟂ BC; hence,

BD = DC

D is the midpoint of BC.

OD bisects ∠BOC

OB = OC (radii)

BD = DC (from Step 2)

O and D lie on perpendicular bisector of BC

Therefore, OD is the perpendicular bisector of chord BC; hence, it bisects the angle at the centre subtended by BC.

∠BOD = ∠COD

Using the Central Angle Theorem

Chord BC subtends:

At center: ∠BOC

At circumference (on ΔABC): ∠A

The angle made at the center is twice the angle made at the circumference by the same chord.

∠BOC = 2∠A     ...[OD bisects ∠BOC]

Consider a ΔABC.

Two circles are drawn while taking AB and AC as the diameter.

Let them intersect each other at D, and let D not lie on BC.

Join AD.

∠ADB = 90°...(Angle subtended by semi-circle)

∠ADC = 90°   ...(Angle subtended by semi-circle)

∠BDC = ∠ADB + ∠ADC = 90° + 90° = 180°

Therefore, BDC is a straight line, and hence, our assumption was wrong.

Thus, point D lies on the third side BC of ΔABC.

Let two circles having their centres as O and O’ intersect each other at point A and B respectively. Let us join OO’.

In ΔAOO’ and BOO’,

OA = OB   ...(Radius of circle 1)

O’A = O’B   ...(Radius of circle 2)

OO’ = OO’   ...(Common)

ΔAOO’ ≅ ΔBOO’   ...(By SSS congruence rule)

∠OAO’ = ∠OBO’   ...(By CPCT)

Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Statement:
If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Short Proof (Idea):

• Given, ∠B + ∠D = 180°.

• Draw a circle through three vertices of the quadrilateral.

• If the fourth vertex does not lie on the circle, an exterior angle becomes equal to its interior opposite angle, which is not possible.

• Hence, the fourth vertex must lie on the same circle.

Conclusion:
Therefore, ABCD is a cyclic quadrilateral.

Statement:

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Short Proof (Idea):

• In a cyclic quadrilateral, the sum of opposite angles is 180°.

• The exterior angle and the adjacent interior angle form a straight line, so their sum is 180°.

• Since both are supplementary to the same angle, they are equal.

Conclusion:

Therefore, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Statement:
The sum of the opposite angles of a cyclic quadrilateral is 180°.

Short Proof (Idea):

• Let ABCD be a cyclic quadrilateral.

• Arc ABC subtends an angle ∠ADC at the circle and ∠AOC at the centre.

• The angle at the centre is double the angle at the circle.

  ∠ADC=`1/2`∠AOC

• Similarly, the other arc subtends:

  ∠ABC = `1/2`(reflex ∠AOC)

• The sum of angles around the centre is 360°.

• Therefore,

  ∠ADC + ∠ABC = `1/2`(360°) = 180^∘

Conclusion:
Hence, the opposite angles of a cyclic quadrilateral are supplementary.

In ΔABC,

∠ABC + ∠BCA + ∠CAB = 180°  ...(Angle sum property of a triangle)

⇒ 90° + ∠BCA + ∠CAB = 180°

⇒ ∠BCA + ∠CAB = 90°    ...(1)

In ΔADC,

∠CDA + ∠ACD + ∠DAC = 180°   ...(Angle sum property of a triangle)

⇒ 90° + ∠ACD + ∠DAC = 180°

⇒ ∠ACD + ∠DAC = 90°   ...(2)

Adding equations (1) and (2), we obtain

∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°

⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°

∠BCD + ∠DAB = 180°     ...(3)

However, it is given that

∠B + ∠D = 90° + 90° = 180°    ...(4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

∠CAD = ∠CBD   ...(Angles in the same segment)

Given: `square`ABCD is a cyclic quadrilateral.

To prove: ∠BAD + ∠BCD = 180º 

∠ABC + ∠ADC = 180º

Proof: Arc BCD is intercepted by the inscribed ∠BAD.

∠BAD = `1/2` m(arc BCD)     ...(i) [Inscribed angle theorem]

Arc BAD is intercepted by the inscribed ∠BCD.

∴ ∠BCD = `1/2` m(arc DAB)      ...(ii) [Inscribed angle theorem]

From (1) and (2) we get

∠BAD + ∠BCD = `1/2` [m(arc BCD) + m(arc DAB)]

∴ (∠BAD + ∠BCD) = `1/2 xx 360^circ`    ...[Completed circle]

= 180°

Again, as the sum of the measures of angles of a quadrilateral is 360°

∴ ∠ADC + ∠ABC = 360° – [∠BAD + ∠BCD]

= 360° – 180°

= 180°

Hence, the opposite angles of a cyclic quadrilateral are supplementary.

It is given that BE is the bisector of ∠B.

∴ ∠ABE = ∠B/2

However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)

⇒ ∠ADE = ∠B/2

Similarly, ∠ACF = ∠ADF = ∠C/2         (Angle in the same segment for chord AF)

∠D = ∠ADE + ∠ADF

`= (angleB)/2 + (angleC)/2`

`= 1/2(angleB + angleC)`

`= 1/2(180^@ - angleA)`

`= 90^@ - 1/2angleA`

Similarly, it can be proved that

`angleE = 90^@ - 1/2angleB`

`angleF = 90^@ - 1/2angleC`

• Diameter → Longest chord of a circle

• Perpendicular from centre to a chord → Bisects the chord

• Line joining centre to midpoint of a chord → Perpendicular to the chord

• The greater the chord → Nearer to the centre

• The smaller the chord, → farther from the centre

• Equal chords → Equidistant from the centre

• Chords equidistant from centre → Equal in length

• Only one circle passes through three non-collinear points