Circles

Tangent is a line that intersects the circle at one point.

Since, there are infinite number of points on circle. At every point, there is one tangent.

Hence, there are infinite number of tangents in a circle.

(i) A tangent to a circle intersects it in One point.

XY is a tangent which intersect the circle at A.

(ii) A line intersecting a circle in two points is called a Secant.

(iii) A circle can have Two parallel tangents at the most.

AB and CD are two parallel tangents

(iv) The common point of a tangent to a circle and the circle is called the Point of contact.

A is point of contact here.

(D)

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

OP Ʇ PQ

By applying Pythagoras theorem in ΔOPQ,

OP² + PQ² = OQ²

5² + PQ² =12²

PQ² =144 - 25

PQ = cm.

Steps of Construction:
1) Draw a circle with any radius and center O.
2) Choose any point P on the circumference of circle, and draw a line passing through P, Let's name it CD.
3) Draw a line AB parallel to CD, such that AB intersects the circle at two points P and A.
Here,
AB and CD are two parallel lines.
AB intersects the circle at exactly two points, P and Q. Therefore, line AB is the secant of this circle.
CD intersects the circle at exactly one point, R, line CD is the tangent to the circle.

(A)

Let O be the centre of the circle.

Given that,

OQ = 25cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact,

Therefore, OP ⊥ PQ

Applying Pythagoras theorem in ΔOPQ, we obtain

OP² + PQ² = OQ²

OP² + 24² = 25²

OP² = 625 - 576

OP² = 49

OP =

Therefore, the radius of the circle is 7 cm.

Given: TP and TQ are tangents.

Therefore, radius drawn to these tangents from centre of the circle will be perpendicular to the tangents.

Thus, OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90°

∠OQT = 90°

In quadrilateral POQT,

Sum of all interior angles = 360°

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360°

90°+ 110° + 90° + ∠PTQ = 360°

∠PTQ = 70°

Given: PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90°

∠OAP = 90°

In quadrilateral AOBP,

Sum of all interior angles = 360°

∠OAP + ∠APB +∠PBO + ∠BOA = 360° 90° + 80° +90° + ∠BOA = 360°

∠BOA = 100°

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

And thus, ∠POB = ∠POA

Let AB is diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively. Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

∠OAR = 90°

∠OAS = 90°

∠OBP = 90°

∠OBQ = 90°

It can be observed that

∠OAR = ∠ OBQ (Alternate interior angles)

∠OAS = ∠ OBP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

to Prove: A line drawn perpendicular from P will pass from O.
Given: APB is tangent to circle with centre O.

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

Assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O'. Join OP and O'P.

As perpendicular to AB at P passes through O', therefore,

∠ O'PB = 90° ... (1)

O is the centre of the circle and P is the point of contact.
We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

∴ ∠ OPB = 90° ... (2)

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

∠ O'PB = ∠ OPB ... (3)

From the figure, it can be observed that,

∠ O'PB < ∠ OPB ... (4)

Therefore, ∠ O'PB = ∠ OPB is not possible. It is only possible, when the line O'P coincides with OP.

Therefore, the perpendicular to AB through P passes through centre O.
Hence, Proved.

Let us consider a circle centred at point O.

AB is a tangent drawn on this circle from point A.

Given that,

OA = 5cm and AB = 4 cm

In ΔABO,

OB ⊥ AB (Radius ⊥ tangent at the point of contact)

Applying Pythagoras theorem in ΔABO, we obtain

AB² + BO² = OA²

4² + BO² = 5²

16 + BO² = 25 BO² = 9

BO = 3

Hence, the radius of the circle is 3 cm.

Let the two concentric circles be centred at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.

OA ⊥ PQ (As OA is the radius of the circle)

Applying Pythagoras theorem in ΔOAP, we obtain

OA² + AP² = OP²

3² + AP² = 5²

9 + AP² = 25

AP² = 16

AP = 4

In ΔOPQ,

Since OA ⊥ PQ,

AP = AQ (Perpendicular from the centre of the circle bisects the chord)

PQ = 2AP = 2 × 4 = 8

Therefore, the length of the chord of the larger circle is 8 cm.

To Prove: AB + CD = AD + BC
Proof:
In the given figure, it can be observed that AB touches the circle at point P; BC touches the circle at point Q; CD touches the circle at point R; DA touches the circle at point S.

Then, we can say from a theorem that " Length of tangents drawn from an external point to the circle are same ",
Therefore,

DR = DS (Tangents on the circle from point D) ….. (1)

CR = CQ (Tangents on the circle from point C) …... (2)

BP = BQ (Tangents on the circle from point B) ….... (3)

AP = AS (Tangents on the circle from point A) ….... (4)

Adding all these equations, we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC
Hence, Proved.

Let us join point O to C.

In ΔOPA and ΔOCA,

OP = OC (Radius of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

ΔOPA ≅ ΔOCA (SSS congruence criterion)

Similarly, ΔOQB ≅ ΔOCB

∠QOB = ∠COB … (ii)

Since POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°

From equations (i) and (ii), it can be observed that 2∠COA + 2 ∠COB = 180°

∠COA + ∠COB = 90°

∠AOB = 90°

To Prove: ∠ APB + ∠ BOA = 180°

Let us consider a circle centred at point O.

Let P be an external point from which two tangents PA and PB are drawn to the circle which touches the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends

∠ AOB at centre O of the circle.

It can be observed that

OA ⊥ PA (radius of circle is always perpendicular to tangent)

Therefore, ∠ OAP = 90°

Similarly, OB ⊥ PB

∠ OBP = 90°

In quadrilateral OAPB,

Sum of all interior angles = 360°

∠ OAP +∠ APB+∠ PBO +∠ BOA = 360°

90° + ∠ APB + 90° + ∠ BOA = 360°

∠ APB + ∠ BOA = 180°

Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Since ABCD is a parallelogram,

AB = CD ...(1)

BC = AD ...(2)

It can be observed that

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain 2AB = 2BC

AB = BC ...(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.

In ∆ABC,

CF = CD = 6cm (Tangents on the circle from point C)

BE = BD = 8cm (Tangents on the circle from point B)

AE = AF = x (Tangents on the circle from point A)

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

2S = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

S =

=√(14 + x)(x)(8)(6)

Area of ΔOBC =

Area of ΔOCA =

Area of ΔOAB =

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

⇒

⇒

⇒

⇒
⇒ 3x = 14 + x

⇒ 2x = 14

⇒ x = 7

Therefore, x = 7

Hence,
AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm
Hence, AB = 15 cm and AC = 13 cm

Let ABCD be a quadrilateral circumscribing a circle centred at O such that it touches the circle at point P, Q, R, S.

join the vertices of the quadrilateral ABCD to the centre of the circle.

Consider ΔOAP and ΔOAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

ΔOAP ≅ ΔOAS (SSS congruence criterion)

And thus, ∠ POA = ∠ AOS

∠1 = ∠ 8 Similarly,

∠2 = ∠ 3

∠4 = ∠ 5

∠6 = ∠ 7

∠1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360°

(∠ 1 + ∠ 8) + (∠ 2 + ∠ 3) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 7) = 360°

2∠ 1 + 2∠ 2 + 2∠ 5 + 2∠ 6 = 360°

2(∠ 1 + ∠ 2) + 2(∠ 5 + ∠ 6) = 360°

(∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180°

∠ AOB + ∠ COD = 180°

Similarly, we can prove that ∠ BOC + ∠ DOA = 180°

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.