Circle

(1) ere CA is the radius of the circle and A is the point of contact of the tangent AB.

⇒ ∠CAB = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.

(2) CA is the radius of the circle which is perpendicular to the tangent AB.

So, the perpendicular distance of line AB from C = CA = 6 cm

(3) In triangle ABC right-angled at A,

Given AB = 6 cm and CA = 6 cm

BC² = AB² + CA² {Using Pythagoras theorem}

⇒ BC² = 6² + 6²

⇒ BC² = 36 + 36

⇒ BC = √72

⇒ BC = 6√2 cm

(4) In triangle ABC right-angled at A,

AB = CA = 6 cm

⇒∠ABC = ∠ACB {Angles opposite to equal sides are equal}

⇒∠ABC + ∠ACB + ∠ BAC = 180° {Angle sum property of the triangle}

⇒ 2∠ABC = 90° {∵ ∠ BAC = 90°}

⇒ ∠ABC = 45°

(1) Here OM is the radius of the circle and M and N are the points of contact of MR and NR respectively.

⇒ ∠RMO = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.

In triangle ORM right-angled at M,

Given that OR = 10 cm and OM = 5 cm {Radius of the circle}

OR² = OM² + RM² {Using Pythagoras theorem}

⇒MR² = 10² -5²

⇒MR² = 100 - 25

⇒ MR = √75

⇒ MR = 5√3 cm

Also, RN = 5√3 cm {∵ Tangents from the same external point are congruent to each other.}

(2)

⇒∠MRO = 30°

(3) Similarly, ∠NRO = 30°

⇒∠MRN = ∠ MRO + ∠NRO = 30° + 30° = 60°

In triangle MOR and triangle NOR,

MR = NR {∵Tangents from same external point are congruent to each other.}

OR = OR {Common}

OM = ON {Radius of the circle}

⇒ ΔMOR ≅ ΔNOR {By SSS}

⇒ ∠ROM = ∠RON

And ∠MRO = ∠NRO {C.P.C.T.}

Hence proved that seg OR bisects ∠MRNas well as ∠MON.

Let BC and DE be the parallel tangents to a circle centered at A with point of contact O and H respectively. On joining OH, we find OH is the diameter of the circle.∠ BOA = 90° = ∠ DHA {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}

Distance between BC and DE = OH

∵ OH is perpendicular to BC and DE.

OH = 2 × 4.5 cm = 9 cm

Given: Two circles are touching each other internally.

∵The distance between the centres of the circles touching internally is equal to the difference of their radii.

⇒distance between their centres = 4.8 cm – 3.5 cm = 1.3 cm

Given: Two circles are touching each other externally

We know that if the circles touch each other externally, distance between their centres is equal to the sum of their radii.

⇒distance between their centres = 5.5 cm + 4.2 cm = 9.7 cm

(i)

Steps of construction:

1. Draw a circle with radius 4cm and centre A.

2. Draw another circle with radius 2.8 cm and centre B such that they touch each other externally.

(ii)

Steps of construction:

1. Draw a circle with radius 4cm and centre A.

2. Draw another circle with radius 2.8 cm and centre B such that they touch each other internally.

(1)In ΔAPR,

AP = RP {Radius of the circle with centre P}

∠PAR = ∠PRA … (1)

In ΔRQB,

RQ = QB {Radius of the circle with centre Q}

∠QRB = ∠QBR …. (2)

⇒ ∠PRA = ∠QRB {Vertically Opposite Angle} ….(3)

⇒ ∠PAR = ∠QBR {From (1), (2) and (3)}

⇒ Alternate interior angles are equal.

⇒ AP || BQ

Hence, proved.

(2) In ∆ APR and ∆ RQB,

∠PAR = ∠QBR and ∠PRA = ∠QRB {From (1) and (2)}

⇒ ∆ APR ~ ∆ RQB {AA}

Hence, proved.

(4) Given: ∠ PAR = 35°

⇒ ∠QBR = 35° = ∠QRB {Proved previously}

In ∆ RQB,

⇒ ∠ RQB + ∠ QRB + ∠QBR = 180° {Angle sum property of the triangle}

⇒∠ RQB + 35° + 35° = 180°

⇒∠ RQB = 180°- 70° = 110°

Given that two circles with centre A and B touch each other externally. We know that if the circles touch each other externally, distance between their centres is equal to the sum of their radii.

⇒AB = (4 + 6) cm = 10 cm

In ∆ABC right-angles at A,

BC² = CA² + AB² {Using Pythagoras theorem}

⇒BC² = 4² + 10²

⇒BC² = 16 + 100

⇒ BC = √116 cm

In ∆DBC,

∠ BDC = 90° because D is the point of contact of tangent CD to circle centred B

BC² = CD² + DB² {Using Pythagoras theorem}

⇒CD² = 116 - 6²

⇒CD² = 116 - 36

⇒ CD = √80 cm = 4√5

Given ∠ECF = 70° and m(arc DGF) = 200°

We know that measure of major arc = 360° - measure of minor arc

m(arc DGF) = 360° - m(arc DF)

⇒ m(arc DF) = 360° - 200° = 160°

⇒∠ DCF = 160°

∵ The measure of a minor arc is the measure of its central angle.

∴m(arc DEF) = 160°

So, ∠DCE = ∠ DCF -∠ECF = 160° - 70°

⇒ ∠DCE = 90°

The measure of a minor arc is the measure of its central angle.

m(arc DE) = 90°

(1) Two arcs are congruent if their measures and radii are equal.

∵∆ QRS is an equilateral triangle

∴ RS = QS = QR

⇒arc RS ≅ arc QS ≅ arc QR

(2) Let O be the centre of the circle.

m(arc QS) = ∠ QOS

∠ QOS + ∠ QOR + ∠ SOR = 360°

⇒ 3∠ QOS = 360° {∵ ∆QRS is an equilateral triangle}

⇒∠ QOS = 120°

m(arc QS) = 120°

m(arc QRS ) = 360° - 120° {∵Measure of a major arc = 360°- measure of its corresponding minor arc}

⇒m(arc QRS ) = 240°

∵ chord AB ≅ chord CD

∴ m(arc AB) = m(arc CD){Corresponding arcs of congruent chords of a circle (or congruent circles) are congruent}

Subtract m(arc CB) from above,

m(arc AB)– m(arc CB) = m(arc CD) – m(arc CB)

⇒m(arc AC) = m(arc BD)

⇒arc AC ≅ arc BD

Hence, proved.

(1)In ∆AOB,

AB = OA = OB = radius of circle

⇒ ∆AOB is an equilateral triangle

∠ AOB + ∠ ABO + ∠ BAO = 180° {Angle sum property}

⇒ 3∠ AOB = 180° {All the angles are equal}

∠ AOB = 60°

(2)∠ AOB = 2 × ∠ ACB {The measure of an inscribed angle is half the measure of the arc intercepted by it.}

⇒∠ ACB = 30°

(3)∠ AOB = 60°

⇒arc(AB) = 60° {The measure of a minor arc is the measure of its central angle.}

(4) Using Measure of a major arc = 360°- measure of its corresponding minor arc

⇒arc(ACB) = 360° - arc(AB)

⇒arc(ACB) = 360° - 60° = 300°

(1) Given PQRS is a cyclic quadrilateral.

∵Opposite angles of a cyclic quadrilateral are supplementary

⇒∠ PSR + ∠ PQR = 180°

⇒∠ PQR = 180° - 110°

⇒∠ PQR = 70°

(2)2 × ∠ PQR = m(arc PR){The measure of an inscribed angle is half the measure of the arc intercepted by it.}

m(arc PR) = 140°

⇒m(arc PQR) = 360° -140° = 220° {Using Measure of a major arc = 360°- measure of its corresponding minor arc}

(3)side PQ ≅ side RQ

∴m(arc PQ) = m(arc RQ){Corresponding arcs of congruent chords of a circle (or congruent circles) are congruent}

⇒m(arc PQR) = m(arc PQ) + m(arc RQ)

⇒m(arc PQR) = 2 × m(arc PQ)

⇒m(arc PQ) = 110°

(4)In ∆ PQR,

∠ PQR + ∠ QRP + ∠ RPQ = 180°{Angle sum property}

⇒∠ PRQ + ∠ RPQ = 180° - ∠ PQR

⇒ 2∠ PRQ = 180° - 70° {∵side PQ ≅ side RQ}

⇒∠ PRQ = 55°

Given MRPN is a cyclic quadrilateral.

⇒∠ R + ∠ N = 180° {Using Opposite angles of a cyclic quadrilateral are supplementary}

⇒ (5x - 13)° + (4x + 4)° = 180°

⇒9x - 9 = 180°

⇒ x – 1 = 20°

⇒ x = 21°

∠R = (5x - 13)° = 5 × 21 – 13 = 105 – 13 = 92°

∠N = (4x + 4)° = 4 × 21 + 4 = 84 + 4 = 88°

Given RS is the diameter

⇒ ∠ ROS = 180°

m(arc RS) = 180°

Now, ∠ RTS is an external angle.

Hence, ∠ RTS is an acute angle.

In ABCD,

∠A = 90°{∵ angle of a rectangle is 90°.}

∠C = 90° {opposite angles are equals}

⇒∠ A + ∠ C = 180°

If opposite angles are supplementary, the quadrilateral is cyclic.

∴ ABCD is cyclic.

(1)In WZPT,

∠ WZP = ∠ WTP = 90° {YZ and XT are the altitudes}

If a pair of opposite angles of a quadrilateral is supplementary, then the

quadrilateral is cyclic.

⇒ WZPT is cyclic.

(2)∵ X, Z,T,Y lie on same circle, ∴ they are concyclic.

Given m(arc NS) = 125°, m(arc EF) = 37°

Also, ∠ NMS is an external angle, so

Given ∠ ABE = 108°, m(arc AE) = 95°

Using the property of the secant,

⇒m(arc DC) = 108° × 2 – 95°

⇒m(arc DC) = 121°

Given PQ = 12, PR = 8

SP × RP = PQ²

This property is known as tangent secant segments theorem.

⇒PS × 8 = 12²

RS = PS – RP = 18 – 8 = 10

(1) Given RD = 15, DS = 4, MD = 8

MD × DN = RD × DS

This property is known as theorem of chords intersecting inside the circle.

⇒ 8 × DN = 15 × 4

(2)Given RS = 18, MD = 9, DN = 8

Here, RS = 18

Let RD = x and DS = 18 – x

MD × DN = RD × DS

This property is known as theorem of chords intersecting inside the circle.

⇒ 8 × 9 = x × (18 – x)

⇒18x – x² = 72

⇒x² – 18x + 72 = 0

⇒ (x – 12)(x – 6) = 0

⇒ x = 12 or 6

⇒ DS = 6 or 12

(1)Given: OE AD, AB = 12, AC = 8

⇒ AD × AC = AB²

This property is known as tangent secant segments theorem.

⇒ AD × 8 = 12²

(2)DC = AD – AC = 18 – 8 = 10

(3)As we know that a perpendicular from centre divides the chord in two equal parts. Here, OE AD.

⇒ DE = EC

⇒ DE + EC = DC

⇒2DE = DC

Given: PQ = 6, QR = 10, PS = 8

PT × PS = PR × PQ

This property is known as theorem of chords intersecting outside the circle.

⇒ PR = PQ + RQ = 6 + 10 = 16

⇒ PT × 8 = 16 × 6

⇒ PT = 12

TS = PT – PS = 12 – 8 = 4

In ∆DEF,

∠DFE = 90° {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}

Given: EF = diameter of the circle.

DE² = DF² + EF² {Using Pythagoras theorem}

⇒ DE² = DF² + (2r)²

⇒ DE² = DF² + 4r²

⇒ DF² = DE²- 4r²

Also, DE × DG = DF²

This property is known as tangentsecant segments theorem.

⇒DE × DG = DE² - 4r²

⇒ DE²- DE × DG = 4r²

⇒DE(DE – DG) = 4r²

⇒ DE × EG = 4r²

Hence, proved.

Given that both the circles touch each other but not specified externally or internally.

The distance between the centres of the circles touching internally is equal to the difference of their radii.

⇒distance between their centres = 5.5 cm – 3.3 cm = 2.2 cm

If the circles touch each other externally, distance between their centres is equal to the sum of their radii.

⇒distance between their centres = 5.5 cm + 3.3 cm = 8.8 cm

Given OA = 12

From the figure, OA is the radius of both the circles.

Given that distance between their centres is OA = 12

∴ Radius of the circles = 12

Let ABCD be a parallelogram which circumscribes the circle.

AP = AS [Tangents drawn from an external point to a circle are equal in length]

BP = BQ [Tangents drawn from an external point to a circle are equal in length]

CR = CQ [Tangents drawn from an external point to a circle are equal in length]

DR = DS [Tangents drawn from an external point to a circle are equal in length]

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

AB + CD = AD + BC

But AB = CD and BC = AD [Opposite sides of parallelogram ABCD]

AB + CD = AD + BC

Hence 2AB = 2BC

Therefore, AB = BC

Similarly, we get AB = DA and DA = CD

Thus, ABCD is a rhombus.

Given: BO = 12.5 cm and AB = 12cm

In ∆AOB,

∠OAB = 90° {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}

BO² = AB² + OA² {Using Pythagoras theorem}

⇒ (12.5)² = 12² + OA²

⇒ OA² = 156.25 - 144

⇒ OA = √12.25

⇒ OA = 3.5cm

Radius = 3.5 cm

⇒ Diameter = 7 cm

If two circles are touching each other externally, they have 3 tangents in common. The above figure proves this statement.

There are three common tangets for the given two circles.

Given ∠ACB = 65°

⇒∠ AOB = 2 × 65° = 130° {∵ The measure of an inscribed angle is half the measure of the arc intercepted by it.}

m(AB) = 130°

So, m(arc ACB ) = 360° - m(AB) {∵ Measure of a major arc = 360°- measure of its corresponding minor arc}

⇒m(arc ACB ) = 360° - 130° = 230°

Given: AE = 5.6,EB = 10, CE = 8

We know that AE × EB = CE × ED

This property is known as theorem of chords intersecting inside the circle.

⇒ 5.6 × 10 = 8 × ED

⇒ ED = 7

Given that 2∠ A = 3∠ C

We know that in a cyclic quadrilateral opposite angles are supplementary to each other.

⇒∠ A + ∠ C = 180°

⇒∠ C = 72°

Angle subtended by the arcs at centre = 120°

⇒ Angle subtended by the arc at the remaining part of the circle = 60° {The measure of an inscribed angle is half the measure of the arc intercepted by it.}

∵ Interior angles of the triangle ABC = 60°

∴It is an equilateral triangle.

If Y would have lied on circumference ∠ XYZ = 90° since XZ is the diameter.

If Y lied outside the circle, ∠ XYZ = acute angle

∴∠ XYZ is an obtuse angle.

Statements (i), (ii) and (iii) are true.

(1)The perpendicular distance of O from P = radius of the circle = 9 cm.

(2)Q lies in the interior of the circle because P lieing on the circumference of the circle is at a distance of 9 cm.

(3)Position of R is not specified.

(1) Here LM is the radius of the circle

⇒ ∠KML = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.

In triangle MLK right-angled at L,

Given MK = 12, KL = 6√3,

MK² = LM² + KL² {Using Pythagoras theorem}

⇒LM² = 12² -6√3²

⇒LM² = 144 - 108

⇒ LM = √36

⇒ LM = 6 cm

(2)

⇒∠K = 30°

∠ M + ∠ K + ∠ L = 180° {Angle sum property of the triangle}

So, ∠ M = 180° - ∠ K - ∠ L

⇒ ∠ M = 180° - 30° - 90° = 60°

Given: AB = r = radius of the circle

Here, AB = AC = r {tangents from the same external point are equal}

And OB = OC = r = radius of the circle.

⇒ ∠ OBA = ∠ OCA = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.

∵ sides of ABOC are equal and opposite angles are 90° each

Hence, ABOC is a square.

Given: AE = 4.5, EB = 5.5

Here, AE = AH = 4.5 {tangents from same external point are equal}

EB = BF = 5.5{tangents from same external point are equal}

∵ Opposite sides of a parallelogram are equal

∴ AE = DG and EB = GC

Also, DH = DG = 4.5{tangents from same external point are equal}

And FC = GC = 5.5 {tangents from same external point are equal}

⇒ AD = AH + HD = 10

(1)MT = radius of the big circle = 9 cm

(2)MN = MT – TN = 9 – 2.5 = 6.5 cm

(3)SM is the tangent to the circle with radius 2.5 cm with S being point of contact.

∠ NSM = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.

In ∆MSN,

∠ MSN = 90°{∵ MS is the tangent to the small circle with point of contact S}

⇒ MN² = MS² + NS²

MS² = MN² – NS²

⇒ MS² = 6.5² – 2.5²

⇒ MS² = 36

⇒ MS = 6 cm

Now, SR = MR – MS = 9 – 6 = 3 cm

⇒ MS:SR = 6:3 = 2:1

Construction: Draw segments XZ and YZ.

Proof: By theorem of touching circles, points X, Z, Y are concyclic.

_{∠XZA =} ∠ YZB {vertically opposite angles}

Let ∠ XZA = ∠ BZY = a (I)

Now, seg XA ≅segXZ (radius of the same circle)

_∵∠XAZ = ∠ XZA = a (isosceles triangle theorem) (II)

Similarly, seg YB ≅ YZ (radius of the same circle)

∴ ∠ BZY = ∠ZBY = a (isosceles triangle theorem) (III)

∴from (I), (II), (III),

∠ XAZ = ∠ ZBY

∴radius XA || radius YB (since alternate interior angles are equal)

XA and YB are the radii of the respective circles.

AZ and BZ are the chords of the circles.

In triangle XAZ,

AX = XZ {Radii of the same circle}

⇒ ∠ XAZ = ∠ XZA {angles opposite to equal sides are equal}

In triangle YBZ,

YB = YZ {Radii of the same circle}

⇒ ∠ YBZ = ∠ YZB {angles opposite to equal sides are equal}

⇒ ∠ XAZ = ∠ XZA = ∠ YBZ = ∠ YZB

∵ Corresponding angles are equal

⇒ XA||YB

The radius of the circle will bisect the chord RS. Therefore, RQ = QS = 1/2 × 12 = 6

Let the radius of circle be r,

Now, in Δ OQS, we have,

RQ = 6

OR = r

OQ = 1/2 r

Applying Pythagoras theorem, we get,

3r² = 4 × 36

r² = 4 × 12 = 48

r = √48 units

To Prove: seg CP ≅seg CQ

Construction: Join CP, CQ and CT

Figure:

Since PQ is a tangent to the circle, ∠CTP = ∠CTQ = 90

Since ∠APT = ∠CTP = 90

AP || CT.

Similarly,

CT || BQ.

So, we can say that,

AP || CT || BQ

AB is a line cutting all three parallel lines.

AC = CB (Radius of the circle, and AB is diameter, C is center)

Since C is center point of line AB cutting parallel lines.

We can say these parallel lines are equal distance.

Therefore, PT = TQ.

Now in ΔCTP and ΔCTQ,

CT is a common side, PT = TQ and ∠CTP = ∠CTQ = 90

CP = CQ. (Pythagoras theorem or congruent triangle theorem)

Hence, Proved.

Draw a circle with radius 3 and centred at A. Similarly draw other two circles with centre B and C and same radius touching each other externally.

We draw a circle of any radius and take any three points A, B and C on the circle.

We join A to B and B to C.

We draw perpendicular bisectors of AB and BC.

We know that perpendicular from the center bisects the chord.

Hence the center lies on both of the perpendicular bisectors.

The point where they intersect is the center of the circle.

The perpendiculars of the line segments drawn by joining collinear points is always parallel whereas in circle any three point’s perpendicular bisector will always intersect at the center.

Hence, any three points on the circle cannot be collinear.

(1) As TAQS is a cyclic quadrilateral,

∠TAQ + ∠TSQ = 180° (Sum of opposite angles of a cyclic quadrilateral is 180° )

(2) ∠ASQ and ∠ATQ

(3) ∠ QAS and ∠SQR

(4) ∠TAS = 65°

∠ TQS = ∠ TAS = 65° (angle by same arc TS in the same sector)

m(arc TS) = ∠TQS + ∠TAS

⇒ m(arc TS) = 65 + 65 = 130°

(5) ∠AQP + ∠AQS + ∠SQR = 180°

⇒ 42 + ∠AQS + 58 = 180

⇒ ∠AQS + 100 = 180

⇒ ∠AQS = 80

∠ AQS + ∠ ATS = 180° (opposite angles of a cyclic quadrilateral)

⇒ 80 + ∠ATS = 180

⇒ ∠ATS = 100°

(1) m(arc PR) = ∠POR = 70°

(2) ∠POQ + ∠ QOS + ∠ROS + ∠POR = 360°

As PQ = RS, ∠ POQ = ∠ROS = 80°

⇒ ∠POQ + ∠ QOS + ∠ROS + ∠POR = 360°

⇒ 80 + ∠QOS + 80 + ∠ 70 = 360

⇒ 230 + ∠ QOS = 360

⇒ ∠QOS = 130°

m(arc QS) = ∠QOS = 130°

(3) m(arc QSR) = ∠QOS + ∠ROS = 130 + 80 = 210°

(1)Given: m(arc WY) = 44°,m(arc ZX) = 68°

We know that

(2)Given: WT = 4.8, TX = 8.0, YT = 6.4

We know that WT × TX = YT × TZ {Using secant-tangent theorem}

⇒ 6.4 × TZ = 4.8 × 8

⇒ TZ = 6

(3)Given: WX = 25, YT = 8,YZ = 26

Let WT = x and TX = 25-x

WT × TX = YT × TZ

⇒x(25-x) = 8 × 26

⇒ (x – 16)(x-9) = 0

⇒ WT = 16 or 9

(1)Given: m(arc CE) = 54°,

m(arc BD) = 23°

∠ CAE is an external angle.

(2)Given: AB = 4.2, BC = 5.4,AE = 12.0

Here, AB × AC = AD × EA

⇒ AD × 12 = 4.2 × 5.4

⇒ AD = 3.36

(3) Given AB = 3.6, AC = 9.0,

AD = 5.4

Here, AB × AC = AD × EA

⇒ AE × 5.4 = 3.6 × 9

⇒ AE = 6

Proof : Draw seg GF.

∠ EFG = ∠ FGH {Alternate interior angles} (I)

∠ EFG = 90°{inscribed angle theorem}(II)

∠ FGH = 90°{inscribed angle theorem} (III)

∴m(arc EG) = 90° from (I), (II), (III).

chord EG ≅ chord FH {Corresponding chords of congruent arcs of a circle (or congruent circles) are congruent}

(1)Given: m(arc PR) = 140°,∠ POR = 36°

∠ ROP is an external angle.

⇒m(arc PQ) = 140° -2 × 36°

⇒m(arc PQ) = 68°

(2)Given: OP = 7.2, OQ = 3.2

Here, RO × OQ = OP²

⇒ RO × 3.2 = 7.2 × 7.2

⇒ RO = 16.2

QR = RO – OQ = 16.2 – 3.2 = 13

(3) Given: OP = 7.2, OR = 16.2

Here, RO × OQ = OP²

⇒ 16.2 × OQ = 7.2 × 7.2

⇒ OQ = 3.2

QR = RO – OQ = 16.2 – 3.2 = 13

We see that the line joining D to E passes through C.

In the smaller circle,

A lies in the semicircle,

∴ ∠EAD = 90°

⇒ DA is perpendicular on the chord EB of the bigger circle.

We know that perpendicular from the center bisects the chord.

Therefore, EA = AB.

Proof: Draw seg OD.

∠ ACB = 90° {angle inscribed in semicircle}

∠ DCB = 45° {CD is the bisector of _}

m(arc DB) = 45° {inscribed angle theorem}

∠ DOB = 90° {definition of measure of an arc} (I)

seg OA ≅seg OB {radii of the circle}(II)

∴line OD is bisector of seg AB From (I) and (II)

∴seg AD ≅seg BD

The figure is shown below:

Draw perpendicular on MN from the center O.

Mark the point as A. Join O to N.

As we know that perpendicular on a chord bisects the chord.

AM = MN/2

⇒ AM = 25/2 = 12.5

Given that LM = 9

⇒ LM + LA = AM

⇒ 9 + LA = 12.5

⇒ LA = 3.5

In Δ OAL,

⇒ OL² = OA² + AL²

⇒ 5² = OA² + (3.5)²

⇒ OA² = 25-12.25

⇒ OA² = 12.75

In Δ OAN,

⇒ ON² = OA² + AN²

⇒ ON² = 12.75 + (12.5)²

⇒ ON² = 12.75 + 156.25

⇒ ON² = 169

⇒ ON = 13

Therefore, the radius of the circle is 13.

We join R to S,

As PQ is the tangent at P, we have

∠RPQ = ∠PSR …………..(1)

As PQ is tangent at Q, we have

∠RQP = ∠RSQ …………………(2)

In ΔRPQ, we have

⇒ ∠RPQ + ∠RQP + ∠ PRQ = 180° (Sum of all angles of a triangle)

⇒ ∠PSR + ∠RSQ + ∠PRQ = 180° (From (1) and (2))

⇒ ∠PSQ + ∠PRQ = 180° (∠PSR + ∠RSQ = ∠PSQ)

Hence Proved.

We join MN.

As PRMN is a cyclic quadrilateral,

∠R + ∠PNM = 180° …………………..(1) (opposite angles of a cyclic quadrilateral)

Also, QSMN is a cyclic quadrilateral,

∠S + ∠ QNM = 180° ……………………(2) (opposite angles of a cyclic quadrilateral)

Adding (1) and (2)

∠ R + ∠S + ∠ PNM + ∠QNM = 360°

⇒ ∠ R + ∠S + 180 = 360 (PQ is a straight line)

⇒ ∠ R + ∠S = 180°

Similarly we have,

As PRMN is a cyclic quadrilateral,

∠P + ∠RMN = 180° …………………..(3) (opposite angles of a cyclic quadrilateral)

Also, QSMN is a cyclic quadrilateral,

∠Q + ∠ SMN = 180° ……………………(4) (opposite angles of a cyclic quadrilateral)

Adding (3) and (4)

∠ P + ∠Q + ∠ RMN + ∠SMN = 360°

⇒ ∠ P + ∠Q + 180 = 360 (RS is a straight line)

⇒ ∠ P + ∠Q = 180°

Therefore, PR ∥ SQ.

We join A to B and A to D and A to E

As BC is a tangent at B, we have

∠ CBD = ∠BAE ………….(1)

As CD is a tangent at D, we have

∠ CDB = ∠DAE ………….(2)

In ΔBCD, we have

⇒ ∠CBD + ∠CDB + ∠ BCD = 180° (Sum of all angles of a triangle)

⇒ ∠BAE + ∠DAE + ∠BCD = 180° (From (1) and (2))

⇒ ∠BAD + ∠BCD = 180° (∠BAE + ∠DAE = ∠BAD)

In quadrilateral ABCD,

We have ∠A + ∠C = 180° (Proved above)

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠B + ∠D + 180 = 360

⇒ ∠B + ∠D = 180

Therefore, opposite angles of the quadrilateral sum to 180. Hence ABCD is a cyclic quadrilateral.

Join D to E , D to F and E to F.

In ΔABE,

⇒ ∠ ABE + ∠BAE + ∠BEA = 180° (Sum of all angles of a triangle)

⇒ ∠ABE + ∠ BAE + 90 = 180

⇒ ∠ABE + ∠ BAE = 90

⇒ ∠ABO + ∠ BAC = 90

⇒ ∠ABO = 90 - ∠ BAC ………………….(1)

In quadrilateral BFOD, we have

We have ∠F = 90 , ∠D = 90

⇒ ∠B + ∠F + ∠O + ∠D = 360°

⇒ ∠B + ∠O + 180 = 360

⇒ ∠B + ∠O = 180

Therefore, BFOD is a cyclic quadrilateral.

∠FBO = ∠FDO (angle by the same arc)

⇒ ∠ABO = ∠FDO

From (1),

⇒ ∠ FDO = 90 - ∠BAC ………………….(2)

In ΔAFC,

⇒ ∠ CAF + ∠FCA + ∠AFC = 180° (Sum of all angles of a triangle)

⇒ ∠CAF + ∠ FCA + 90 = 180

⇒ ∠CAF + ∠ FCA = 90

⇒ ∠BAC + ∠ OCE = 90

⇒ ∠OCE = 90 - ∠ BAC ………………….(3)

In quadrilateral CEOD, we have

We have ∠E = 90 , ∠D = 90

⇒ ∠C + ∠E + ∠O + ∠D = 360°

⇒ ∠C + ∠O + 180 = 360

⇒ ∠C + ∠O = 180

Therefore, CEOD is a cyclic quadrilateral.

∠ODE = ∠OCE (angle by the same arc)

From (3),

⇒ ∠ ODE = 90 - ∠BAC …………(4)

From (2) and (4) we conclude,

∠FDO = ∠ODE

OD bisects ∠D.

Similarly, we can prove that OE bisects ∠E and OF bisects ∠F.

Hence O is the incenter of ΔDEF.