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Solution: Smallest number divisible by 4 after 10 is 12, The greatest number below 250 which is divisible by 4 is 248
Number of terms: {(248-12)/4}+1
{236/4}+1 = 59+1 = 60

Solution: a = 8, and n = 15

s₁₅ = 15/2(16+8 x14) =15 x 64 =960

Solution: a = 8 , d = 8 and n = 15

S₁₅ = 15/2(16 +8 x 14) = 15 x 64 =960

Circles with centres R and S lie in the same plane and intersect with a line l in the plane in one and only one point T [R - T - S].

Hence the given circles are externally touching circles.

Solution: a + 3d + a + 7d = 24
Or, 2a + 10d = 24
Similarly, 2a + 14d = 44
So, 44 – 24 = 4d
Or, d = 5
2a + 10x5 = 24
Or, a + 25 = 12
Or, a = -13

So, first three terms of AP: -13, -8, -3,

Solution: 7000 = 5000 + 200(n-1)
Or, 200(n-1) = 2000
Or, n-1 = 10
Or, n = 11

Correct answer is (b) 10 cm

Since the tangents from an external point are equal, we have

AD = AE, CD = CF, BE = BF

Perimeter of △ABC = AC + AB + CB

= (AD - CD) + (CF + BF) + (AE - BE)

= (AD - CF) + (CF + BF) + (AE - BF)

= AD + AE

= 2AE

= 2 x 5

= 10 cm

Proof:

i. ∠DAB ≅ ∠CAB [Given] 

∴ ∠SAM ≅ ∠TAM (i) [A – S – D, A – M – B, A -T – C] 

In ∆SAM and ∆TAM, 

∠SAM ≅ ∠TAM [From (i)] 

∠ASM ≅ ∠ATM [Each angle is of measure 90°] 

seg AM ≅ seg AM [Common side]

∴ ∆SAM ≅ ∆TAM [AAS [SAA] test of congruency] 

∴ side MS ≅ side MT [c.s.c.t] 

But, seg MS ⊥ chord AD [Given] 

seg MT ⊥chord AC

∴ chord AD ≅ chord AC [Chords of a circle equidistant from the centre are congruent]

ii. Theorem used for solving the problem: 

The chords which are equidistant from the centre are equal in length. 

iii. Test of congruency useful in solving the above problem is AAS ISAAI test of congruency.

i. Line AB is the tangent at point B and seg AD is the secant. [Given] 

∴ AC × AD = AB² [Tangent secant segments theorem] 

∴ 8 × AD = 122 

∴ 8 × AD = 144

∴ AD = 144/8

∴ AD = 18 units 

ii. AD = AC + DC [A – C – D] 

∴ 18 = 8 + DC 

∴ DC = 18 – 8 

∴ DC = 10 units 

iii. seg OE ⊥ seg AD [Given] 

i.e. seg OE ⊥ seg CD [A – C – D] 

∴ DE = (1/2) DC [Perpendicular drawn from the centre of the circle to the chord bisects the chord] 

= (1/2) × 10 

∴ DE = 5 units

Consider △ OAB and △ OAC

It is given that AB = AC

OA is common i.e. OA = OA

From the figure we know that OB = OC which is the radii

By SSS congruence criterion

△ OAB ≅ △ OAC

∠OAB = ∠OAC (c. p. c. t)

Therefore, it is proved that O lies on the bisector of ∠BAC.

Proof: 

In ∆OMR and ∆ONR, 

seg RM ≅ seg RN [Tangent segment theorem] 

seg OM ≅ seg ON [Radii of the same circle] 

seg OR ≅ seg OR [Common side] 

∴ ∆OMR ≅ ∆ONR [SSS test of congruency] 

{∴ ∠MRO ≅ ∠NRO ∠MOR ≅ ∠NOR } [c.a.c.t.] 

∴ seg OR bisects ∠MRN and ∠MON.

Given: C is the centre of circle. seg AB is the diameter of circle. line PQ is a tangent, seg AP ⊥ line PQ and seg BQ ⊥ line PQ. 

To prove: seg CP ≅ seg CQ 

Construction: Draw seg CT, seg CP and seg CQ. 

Proof: Line PQ is the tangent to the circle at point T. [Given] 

∴ seg CT ⊥ line PQ (i) [Tangent theorem] 

Also, seg AP ⊥ line PQ, seg BQ ⊥ line PQ [Given] 

∴ seg AP || seg CT || 

seg BQ [Lines perpendicular to the same line are parallel to each other] 

∴ AC/CB = PT/TQ [Property of intercepts made by three parallel lines and their transversals] 

But, AC = CB [Radii of the same circle] 

∴ = AC/AC = PT/TQ

∴ PR/TQ = 1 

∴ PT = TQ ………… (ii) 

∴ In ∆CTP and ∆CTQ, 

 seg PT ≅ seg QT [From (ii)] 

∠CTP ≅ ∠CTQ [From (i), each angle is of measure 90° ] 

seg CT ≅ seg CT [Common side] 

∴ ∆CTP ≅ ∆CTQ [SAS test of congruence] 

∴ seg CP ≅ seg CQ [c.s.c.t]

Proof:

In ∆OMR and ∆ONR,

seg RM ≅ seg RN [Tangent segment theorem]

seg OM ≅ seg ON [Radii of the same circle]

seg OR ≅ seg OR [Common side]

∴ ∆OMR ≅ ∆ONR [SSS test of congruency]

{∴ ∠MRO ≅ ∠NRO

∠MOR ≅ ∠NOR } [c.a.c.t.]

∴ seg OR bisects ∠MRN and ∠MON.

i. Line KL is the tangent to the circle at point L and seg ML is the radius. [Given] 

∴ ∠MLK = 90°…………. (i) [Tangent theorem] 

In ∆MLK, ∠MLK = 90°

∴ MK² = ML² + KL² [Pythagoras theorem] 

∴ 12² = ML² + (6√3)² 

∴ 144 = ML² + 108 

∴ ML² = 144 – 108 

∴ ML² = 36 

∴ ML = √36 = 6 units. [Taking square root of both sides]

∴ Radius of the circle is 6 units.

ii. We know that, 

ML = 1/2 MK 

∴ ∠K = 30° …………… (ii) [Converse of 30° – 60° – 90° theorem] 

In ∆MLK , 

∠L = 90° [From (i)] 

∠K = 30° [From (ii)] 

∴ ∠M = 60° [Remaining angle of ∆MLK]

Chords AC and DE intersect internally at point B.

∴ ∠ABE = 1/2 [m(arc AE) + m(arc DC)]

∴ 108° = 1/2 [95° + m(arc DC)]

∴ 108° × 2 = 95° + m(arc DC)

∴ 95° + m(arc DC) = 216° 

∴ m(arc DC) = 216° – 95° 

∴ m(arc DC) = 121°

Since angles on the same segment of a circle are equal 

Therefore, 

∠CAD = ∠DBC = 55° 

∠DAB = ∠CAD + ∠BAC 

= 55° + 45° = 100° 

But, 

∠DAB + ∠BCD = 180° 

(Opposite angles of a cyclic quadrilateral) 

Therefore, 

∠BCD = 180° – 100° 

∠BCD = 80°

‘O’ is the centre of the circle. 

∴ In ΔOAB; OA = OB (radii) 

∴ ∠OAB = ∠OBA = 40° 

(∵ angles opp. to equal sides) 

Now ∠AOB = 180° – (40° + 40°)

 (∵ angle sum property of ΔOAB) 

= 180°-80° = 100°

But ∠AOB = ∠COD = 100° 

Also ∠OCD = ∠ODC [OC = OD] 

= 40° as in ΔOAB 

∴ ∠BCD = 40° 

(OR) 

In ΔOAB and ΔOCD 

OA = OD (radii) 

OB = OC (radii) 

∠AOB = ∠COD (vertically opp. angles) 

∴ ΔOAB ≅ ΔOCD 

∴ ∠BCD = ∠OBA = 40° 

[ ∵ OB = OA ⇒ ∠DAB = ∠DBA]

(c) 90°

∠CDB =\(\frac{1}{2}\) (180º - 44º) = \(\frac{1}{2}\) x 136º = 68º    (∵ BCD is an isos. Δ)

∠BAD = 180° – 44° = 136° (opp.∠s of a cyclic quad. are supp.)

∴ ∠ADB = \(\frac{1}{2}\) (180º -136º) = \(\frac{1}{2}\) x 44º = 22º  (∵BAD is an isos. Δ)

∴ ∠ADC = ∠ADB + ∠BDC = 22° + 68° = 90°

\(\Rightarrow\) x = ∠ADE = 180°–∠ADC =180° – 90° = 90° (∵ EDC is a st. line)

Given:

PB = 10 cm

CQ = 2 cm

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Using the above property,

PA = PB = 10 cm (tangent from P)

And,

CA = CQ= 10 cm (tangent from C)

Now,

PC = PA – CA

= 10 cm – 2 cm

= 8 cm

Hence, PC = 8 cm

Answer is B. 30°

Given: 

∠TQP = 60° 

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

By property 1, 

TP = TQ (tangent from T) 

⇒ ∠TPQ = ∠TQP = 60° 

By property 2, ∆OPT is right-angled at ∠OPT (i.e., ∠OPT = 90°) and ∆OQT is right-angled at ∠OQT (i.e., ∠OQT = 90°). 

Now, 

∠OPQ = ∠OPT – ∠TPQ 

⇒ ∠OPQ = 90° – 60° 

⇒ ∠OPQ = 30° 

Hence, ∠OPQ = 30°

Given:

PB = 10 cm 

CQ = 2 cm

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Using the above property, 

PA = PB = 10 cm (tangent from P) 

DB = DQ= 10 cm (tangent from D) 

And, 

CA = CQ= 10 cm (tangent from C) 

Now,

Perimeter of ∆PCD = PC + CD + DP

= PC + CQ + QD + DP 

= PC + CA + DB + PD [∵CA = CQ and DB = DQ] 

= PA + PB [∵PA = PC + CA and PB = PD + BD] 

= 10 cm + 10 cm 

= 20 cm 

Hence, Perimeter of ∆PCD = 20 cm

Answer is C. 30°

Given: 

APB = 30° 

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

Property 2: Sum of all angles of a triangle = 180° 

By property 1, 

PA = PB (tangent from P) 

And, 

∠PAB = ∠PBA [∵PA = PB] 

By property 2, 

∠PAB + ∠PBA + ∠APB = 180° 

⇒ ∠PAB + ∠PBA + 30° = 180° 

⇒ ∠PAB + ∠PBA = 180° - 30° 

⇒ ∠ PAB + ∠ PBA = 150° 

⇒ ∠ PBA + ∠ PBA = 150° [∵∠PAB = ∠PBA] 

⇒ 2∠PBA = 150°

⇒ ∠PBA = \(\frac{150^°}{2}\)

⇒ ∠PBA = 75° 

Now, 

∠PBA = ∠CAB = 75° [Alternate angles] 

∠PBA = ∠ACB = 75° [Alternate segment theorem] 

Again by property 2, 

∠CAB + ∠ACB + ∠CBA = 180° 

⇒ 75° + 75° + ∠CBA = 180°

⇒ 150° + ∠CBA = 180° 

⇒ ∠CBA = 180° - 150° 

⇒ ∠CBA = 30° 

Hence, ∠CBA = 30°

Correct answer is (D) 55°

A circle is a collection of all such points in a plane which are equidistant from a fixed point. This fixed point is called the centre of the circle while the distance of any point on the circle from the centre is called the radius of the circle.

Angle between the radii = 180°- 30° = 150°

(Since the sum of opposite angles = 180°)

Answer is A. 18 cm

Given: 

EK = 9 cm

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By above property, 

EM = EK = 9 cm (tangent from E) 

DK = DH (tangent from D) 

FM = FH (tangent from F) 

Now, 

Perimeter of ∆EDF = ED + DF + FE 

⇒ Perimeter of ∆EDF = (EK – KD) + (DH + HF) + (EM – MF) [∵ED = EK – KD 

DF = DH + HF 

FE = EM – MF] 

⇒ Perimeter of ∆EDF = EK – KD + DH + HF + EM – MF 

⇒ Perimeter of ∆EDF = EK – DH + DH + HF + EM – HF [∵DK = DH and FM = FH] 

⇒ Perimeter of ∆EDF = EK + EM 

⇒ Perimeter of ∆EDF = 9 cm + 9 cm 

⇒ Perimeter of ∆EDF = 18 cm 

Hence, Perimeter of ∆EDF = 18 cm

If a circle with any measurement, to find out centre of circle, steps are as follows : 

1. Construct a circle with any radius. 

2. Draw two chords AB and CD with any length. 

3. Draw perpendicular bisector for AB and CD chords. 

4. If these bisectors intersect each other when produced they meet at ’O’. O is the centre of the circle.

Data: OB is the radius, BA is the tangent.

∴ ∠OBA = 90°, OA = 5 cm. (data)

Length of tangent, AB = 4 cm.

∴ Radius, OB = ?

In ⊥∆OBA, ∠OBA = 90°

∴ OB² + BA² = OA². OB² + (4)² = (5)²

OB² + 16 = 25

OB² = 25 – 16

OB² = 9

∴ OB = 3 cm.

∴ Radius of circle, OB = 3cm.

Given equation of circle is 

x² + y² + 2x – 2y – 3 = 0 ….(i) 

Given equation of line is 2x + y + 6 = 0 y = -6 – 2x ……(ii) 

Substituting y = -6 – 2x in (i), we get 

x + (-6 – 2x)² + 2x – 2(-6 – 2x) – 3 = 0

⇒ x² + 36 + 24x + 4x² + 2x + 12 + 4x – 3 = 0 

⇒ 5x² + 30x + 45 = 0 

⇒ x² + 6x + 9 = 0 

⇒ (x + 3)² = 0

⇒ x = -3 

Since, the roots are equal. 

∴ 2x + y + 6 = 0 is a tangent to x² + y² + 2x – 2y – 3 = 0 

Substituting x = -3 in (ii), we get 

y = -6 – 2(-3) = -6 + 6 = 0 

Point of contact = (-3, 0)

Data: Perpendicular at the point of contact to the tangent to a circle passes through the centre. 

PQ is the tangent to circle with centre ’O’. 

Let the perpendicular drawn at the point P is 

∠RPQ. ∠RPQ = 90° …………. (i) 

Radius drawn to circle at the point of contact is perpendicular. 

∴ ∠OPQ = 90° …………. (ii) 

From eqn. (i) and eqn. (ii), we have 

∠RPQ = ∠OPQ = 90° . 

This is contradiction, because 

∠RPQ is the part of ∠OPQ. 

∴ ∠RPQ < ∠OPQ 

∴ ∠RPQ ≠∠OPQ. 

∴ The perpendicular at the point of contact to the tangent to a circle passes through the centre.

(a) Point circle means circle whose radius = 0 

Given C = (4,-5) & r = 0 

The point circle & (x – 4)² + (y + 5)² = 0 

⇒ x² + y² – 8x + 10y + 41 = 0 

(b) Centre (-3,2) 

The point circle is (x + 3)² + (y – 2)² = 0 

⇒ x² + y² + 6x – 4y + 13 = 0 

(C) Centre (1,0) 

∴ Circle is (x – 1)² + (y – 0)² = 0 

x² – 2x + 1 + y² = 0 

⇒ x² + y² – 2x + 1 = 0.