Category: OP Malhotra Solutions

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S Chand Class 11 ICSE Maths Solutions Chapter 18 Limits Ex 18(i)

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S Chand Class 11 ICSE Maths Solutions Chapter 18 Limits Ex 18(h)

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S Chand Class 11 ICSE Maths Solutions Chapter 18 Limits Ex 18(g)

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S Chand Class 11 ICSE Maths Solutions Chapter 18 Limits Ex 18(f)

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S Chand Class 11 ICSE Maths Solutions Chapter 18 Limits Ex 18(c)

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Continuous practice using S Chand ISC Maths Class 11 Solutions Chapter 20 Measures of Central Tendency Chapter Test can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 20 Measures of Central Tendency Chapter Test

Question 1.
The weights of 50 apples were recorded as given below. Calculate the mean weight to the nearest gram, by step deviation method.

Weight in grams	80-85	85-90	90-95	95-100	100-105	105-110	110-115
Number of apples	5	8	10	12	8	4	3

Solution:

Weight (in gms)	No. of apples (frequency fi)	Mid-Marks xi	di = xi – 97.5 A = 97.5	ui = \(\frac{d_i}{i}\) i = 5	fiui
80-85	5	82.5	– 15	-3	– 15
85-90	8	87.5	– 10	-2	– 16
90-95	10	92.5	-5	– 1	– 10
95-100	12	97.5	0	0	0
101 – 105	8	102.5	5	1	8
105-110	4	107.5	10	2	8
110-115	3	112.5	15	3	9
	Σfi = 50				Σfiui = – 16

Then by step deviation method \(\bar{x}\) = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × i
⇒ \(\bar{x}\) = 97.5 + \(\frac{(-16)}{50}\) × 5 = 97.5 – 1.6 = 95.9
Thus the required mean weight be 95.9 grams.

Question 2.
Find the value of p if the mean of the following distribution is 7.5 :

X	2-4	4-6	6-8	8-10	10-12	12-14
f	6	8	15	P	8	4

Solution:
The table of values is given as under:

Class	frequency (fi)	Mid-Marks xi	fixi
2-4	6	3	18
4-6	8	5	40
6-8	15	7	105
8-10	P	9	9P
10-12	8	11	88
12-14	4	13	52
	Σfi =41 +P		Σfixi = 303 + 9p

Then by direct method, Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) ⇒ 7.5 = \(\frac{303+9 p}{41+p}\)
⇒ 7.5 (41 + p) = 303 + 9p ⇒ 1.5p = 4.5 ⇒ p = 3

Question 3.
The mean of the following frequency distribution is 57.6 and the number of observations is 50. Find the missing frequencies f₁ and f₂.

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	7	A	12	A	8	5

Solution:
We construct the table of values is given as under :

=

=

Classes	Class Mark xi	Frequency fi	di = xi – A A = 50	ui = \(\frac{d_i}{C}\) C = 20	fiui
0-20	10	7	-40	-2	– 14
20-40	30	f1	-20	– 1	-h
40-60	50	12	0	0	0
60-80	70	f2	20	1	h
80-100	90	8	40	2	16
100-120	110	5	60	3	15
		Σfi = 32 + f1+f2			Σfiui = 17 – f1+f2

since sum of all observations = 50
∴ Σf_i = 50
⇒ 50 = 32 + f₁ + f₂
⇒ f₁ + f₂ = 18
By step deviation method, we have
Mean = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × C
⇒ 57.6 = 50 + \(\frac{17-f_1+f_2}{32+f_1+f_2}\) × 20
⇒ 57.6 – 50 = \(\frac{17-f_1+f_2}{50}\) × 20
⇒ 5 × 7.6 = 2 (17 – f₁ + f₂)
⇒ 38 = 34 – 2f₁ + 2f₂
⇒ 4 = -2f₁ + 2f₂
⇒ -f₁ + f₂ = 2 …(2)
On adding eqn. (1) and eqn. (2); we have
2f₂ = 20 ⇒ f₂ = 10
∴ from(1); f₁ = 18 – 10 = 8

Effective S Chand ISC Maths Class 11 Solutions Chapter 20 Measures of Central Tendency Ex 20(a) can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 20 Measures of Central Tendency Ex 20(a)

Question 1.
Find the mean of :
(i) the first 6 natural numbers.
(ii) the first ten odd natural numbers.
(iii) the first eight even natural numbers.
(iv) x, x + 2, x + 3, x + 6 and x + 9.
(v) Squares of the first n natural numbers.
(vi) Squares of the first 10 natural numbers.
(vii) Cubes of first n even natural numbers.
Solution:
(i) First six natural numbers are ; 1, 2, 3, 4, 5, 6
∴ required mean = \(\frac{1+2+3+4+5+6}{6}\) = \(\frac{21}{6}\) = \(\frac{7}{2}\)

(ii) First ten add natural numbers are = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
∴ required mean = \(\frac{Sum of all observations}{No. of observations}\) = \(\frac{1+3+5+7+9+11+13+15+17+19}{10}\) = \(\frac{100}{10}\) = 10

(iii) First eight even natural numbers are ; 2, 4, 6, 8, 10, 12, 14, 16
∴ required mean = \(\frac{2+4+6+8+10+12+14+16}{8}\) = \(\frac{72}{8}\) = 9

(iv) Required mean = \(\frac{Sum of all observations}{No. of observations}\) = \(\frac{x+x+2+x+3+x+6+x+9}{5}\) = \(\frac{5 x+20}{5}\) = \(\frac{5(x+4)}{5}\) = x + 4

(v) Squares of first n natural numbers are ; 1², 2², 3²,…, n²
∴ required mean = \(\frac{1^2+2^2+\ldots+n^2}{n}\) = \(\frac{\sum n^2}{n}\) = \(\frac{n(n+1)(2 n+1)}{6 n}\) = \(\frac{(n+1)(2 n+1)}{6}\)

(vi) Squares of first 10 natural numbers are ; 1², 2², 3²,…, 10²

(vii) Class of first n even natural numbers are 2³, 4³, 6³,…, (2n)³
∴ required mean = \(\frac{2^3+4^3+6^3+\ldots+(2 n)^3}{n}\) = \(\frac{2^3\left[1^3+2^3+3^3+\ldots .+n^3\right]}{n}\) = \(\frac{8 \Sigma n^3}{n}\) = \(\frac{8 n^2(n+1)^2}{4 n}\) = 2n(n + 1)²

Question 2.
(i) Find the mean of the following sets of numbers :
(a) 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6
(b) -6, -2, -1, 0, 1, 2, 5, 9.
(ii) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
Solution:
(i) (a) Required mean = \(\frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7}\) = \(\frac{21}{7}\) = 3
(b) Required Mean = \(\frac{(-6)+(-2)+(-1)+0+1+2+5+9}{8}\) = \(\frac{8}{8}\) = 1
(ii) Q. 4 on P-744 (understanding)

Question 3.
Duration of sunshine (in hours) in Delhi for first ten days of month as reported by Meteorological department are as under:
5.1, 4.7, 6.1, 1.7, 5.4, 5.0, 11.7, 11.6, 11.9, 3.2
(i) Calculate \(\bar{x}\);
(ii) Check \(\sum_{i=1}^{10}\left(x_i-\bar{x}\right)\) = 0.
Solution:
Given observations are ; 5.1, 4.7, 6.1, 1.7, 5.4, 5.0, 11.7, 11.6, 11.9, 3.2
Here n = 10
(i) ∴ \(\bar{x}\) = \(\frac{\Sigma x}{n}\) = \(\frac{5.1+4.7+6.1+1.7+5.4+5.0+11.7+11.6+11.9+3.2}{10}\) = \(\frac{332}{50}\) = 6.64
(ii) \(\sum_{i=1}^{10}\left(x_i-\bar{x}\right)\) = \(\sum_{i=1}^{10} x_i-\sum_{i=1}^{10} \bar{x}\) = (x₁ + x₂ + ….. + x₁₀) = 10\(\bar{x}\) = 66.4 – 10 × 6.64 = 66.4 – 66.4 = 0

Question 4.
M being the mean of x₁, x₂, x₃, x₄, x₅ and x₆. Find the value of \(\sum_{i=1}^6\left(x_i-M\right)\).
Solution:
Given observations are x₁, x₂, x₃, x₄, x₅ and x₆
∴ Mean = M = \(\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\) ⇒ x₁ + x₂ + x₃ + x₄ + x₅ + x₆ 6M …(1)
∴ \(\sum_{i=1}^6\left(x_i-\mathrm{M}\right)\) = (x₁ – M) + (x₂ – M) + (x₃ – M) + (x₄ – M) + (x₅ – M) + (x₆ – M)
= (x₁ + x₂ + x₃ + x₄ + x₅ + x₆) – 6M = 6M – 6M = 0

Question 5.
Find the mean of the following frequency distributions :
(i)

X	19	21	23	25	27	29	31
f	13	15	16	18	16	15	13

(ii)

X	2.5	7.5	12.5	17.5	22.5
f	4	5	7	12	7

Solution:
(i) The table of values is given as under :

x	f	d = x – 25 A = 25	fd
19	13	-6	-68
21	15	-4	-60
23	16	-2	-32
25	18	0	0
27	16	2	32
29	15	4	50
31	13	6	78
	N = Σf = 106		Σfd = 0

Then by short-cut method, we have required mean = \(\bar{x}\) = A + \(\frac{\Sigma f d}{\Sigma f}\) = 25 + \(\frac{0}{106}\) = 25

(ii) The table of values is given as under :

x	f	d = x – A A = 12.5	fd
2.5	4	-10	-40
7.5	5	-5	-25
12.5	7	0	0
17.5	12	5	60
22.5	7	10	70
27.5	5	15	75
	Σf = 40		Σfd = 140

Then by short-cut method, we have required Mean = \(\bar{x}\) = \(\frac{\Sigma f d}{\Sigma f}\) = 12.5 + \(\frac { 140 }{ 40 }\) = 12.5 + 3.5 = 16

Question 6.
Calculate the mean of the following data by short cut method :

x	2.5	7.5	12.5	17.5	22.5
f	4	5	7	12	7

Solution:
The table of values is given as under :

x	f	d = x – A A = 12.5	fd
2.5	4	-10	-40
7.5	5	-5	-25
12.5	7	0	0
17.5	12	5	60
22.5	7	10	70
27.5	5	15	75
Σf = 40			Σfd = 140

Then by short-cut method, we have
required Mean = \(\bar{x}\) = A + \(\frac{\Sigma f d}{\Sigma f}\) = 12.5 + \(\frac{140}{40}\) = 12.5 + 3.5 = 16

Question 7.
Calculate by step-deviation method, the arithmetic mean of the following marks obtained by students in English.

Marks	5	10	15	20	25	30	35	40	45	50
No. of students	20	43	75	67	72	45	39	9	8	6

Solution:
The table of values is given as under :

X	f	d = x – A A = 25	Ui = d/i i = 5	fu
5	20	-20	-4	-80
10	43	-15	-3	– 129
15	75	-10	-2	– 150
20	67	-5	– 1	-67
25	72	0	0	0
30	45	5	1	45
35	39	10	2	78
40	9	15	3	27
45	8	20	4	32
50	6	25	5	30
	Σf = 384			Σfu = – 214

Then by step deviation method, we have Mean \(\bar{x}\) = A + \(\frac{\Sigma f u}{\Sigma f}\) × i = 25 – \(\frac{214}{384}\) × 5 = 22.21

Question 8.
The frequency distribution of marks obtained by 40 students of a class is as under. Calculate the Arithmetic Mean.

Marks	0-8	8-16	16-24	24-32	32-40	40-48
Students	6	3	10	16	4	2

Solution:
The table of values is given as under :

Marks	Frequency fi	Class Marks xi	di= Xi – A A = 20	Ui = di/8 i = 8	fiui
0-8	5	4	– 16	-2	– 10
8-16	3	12	-8	– 1	-3
16-24	10	20	0	0	0
24-32	16	28	8	1	16
32-40	4	36	16	2	8
40-48	2	44	24	3	6
	Σfi = 40				Σfiui = 17

Then by step deviation method,
Mean \(\bar{x}\) = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × i = 20 + \(\frac{17}{40}\) × 8 = 20 + 3.4 = 23.4

Question 9.
Compute the mean of the following frequency table by :
(i) direct method and
(ii) short-cut method

Class	Frequency	Class	Frequency
5-10	10	30-35	4
10-15	6	35-40	2
15-20	4	45-45	1
20-25	12	45-50	3
25-30	8

Solution:
The table of values is given as under :

Class	frequency fi	Mid Marks xi	fixi	di = Xi – A A = 27.5	ui = \(\frac{d_i}{i}\) i = 5	fidi
5-10	10	7.5	75	-20	-4	-200
10-15	6	12.5	75	– 15	-3	-90
15-20	4	17.5	70	– 10	-2	-40
20-25	12	22.5	270	-5	– 1	-60
25-30	8	27.5	220	0	0	0
30-35	4	32.5	130	5	1	20
35-40	2	37.5	75	10	2	20
40-45	1	42.5	42.5	15	3	15
45-50	3	47.5	142.5	20	4	60
	Σfi = 50		Σfixi = 1100			Σfidi = – 275

(i) By direct method \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = 22
(ii) By short cut method, \(\bar{x}\) = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\) = 27.5 – \(\frac { 275 }{ 50 }\) = 27.5 – 5.5 ⇒ \(\bar{x}\) = 22

Question 10.
In a city, the following weekly observations were made in a study of cost of living index for year 1970-71.

Cost of living index	140-150	150-160	160-170	170-180	180-190	190 – 200
Number of weeks	5	10	18	9	6	4

(i) Calculate the average weekly cost of living index.
(ii) Verify Σf_i (x_i – \(\bar{x}\)) = 0, where x is the mid-value.
Solution:
The table of values is given as under :

Cost of living Index	No. of weeks fi	Mid-Marks xi	fixi	xi – \(\bar{x}\)	fi (xi – \(\bar{x}\))
140-150	5	145	725	-22.5	– 112.5
150-160	10	155	1550	-12.5	– 125
160- 170	18	165	2970	-2.5	-45
170-180	9	175	1575	7.5	67.5
180-190	6	185	1110	17.5	105
190-200	4	195	780	27.5	110
	Σfi = 52		Σfixi = 8710		Σfi (xi – \(\bar{x}\)) = 0

(i) ∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac { 8710 }{ 52 }\) = 167.5
(ii) ∴Σf_i (x_i – \(\bar{x}\)) = 0

Question 11.
Calculate the mean by step-deviation method for the following data :

Height (in cm)	No. of boys	Height (in cm)	No. of boys
135-140	4	155-160	24
140 -145	9	160-165	10
145-150	18	165-170	5
150-155	28	170-175	2

Solution:
The table of values is given as under :

Height (in cm)	No. of boys fi	Mid-Marks xi	di = xi – A A = 152.5	ui = \(\frac{d_i}{i}\) i = 5	fiui
135-140	4	137.5	– 15	-3	– 12
140-145	9	142.5	– 10	-2	– 18
145-150	18	147.5	-5	– 1	– 18
150-155	28	152.5	0	0	0
155-160	24	157.5	5	1	24
160-165	10	162.5	10	2	20
165-170	5	167.5	15	3	15
170-175	2	172.5	20	4	8
	Σfi=100				Σfiui = 19

Then by step deviation method, we have
\(\bar{x}\) = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × i = 152.5 + \(\frac { 19 }{ 100 }\) × 5 = 152.5 + 0.95 = 153.45

Question 12.
Given:

Variable	20	19	18	17	16	15	14	13	12	11
Frequency	1	2	4	8	11	10	7	4	2	1

Find the mean variable by taking 11 as the assumed mean, and verify by the direct method.
Solution:
The table of values is given as under:

Variable xi	Frequency fi	fixi	di = xi – A A = 15	fidi
20	1	20	5	5
19	2	38	4	8
18	4	72	3	12
17	8	136	2	16
16	11	176	1	11
15	10	150	0	0
14	7	98	– 1	-7
13	4	52	-2	-8
12	2	24	-3	-6
11	1	11	-4	-4
	Σfi = 50	Σfixi = 777		Σfidi = 27

Then by direct method, \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{777}{50}\) = 15.54
By short at method, mean \(\bar{x}\) = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\) = 15 + \(\frac{27}{50}\) = 15.54

Question 13.
The following table shows the distribution of orders of a firm, according to their value :

Value	Under ?10	? 10 and Under ?20	? 20 and Under ?30	? 30 and Under ?40	? 40 and Under ? 50	? 50 and Under ? 60	? 60 and Under ? 70
No. of buyers	245	383	205	89	47	21	10

Estimate (i) the total turnover, (ii) the mean value of a single order.
Solution:
The table of values is given as under :

Classes	No. of buyers fi	Mid-value xi	fixi
0-10	245	5	1225
10-20	383	15	5745
20-30	205	25	5125
30-40	89	35	3115
40-50	47	45	2115
50-60	21	55	1155
60-70	10	65	650
	Σfi = 1000		Σfixi = 19140

∴ required turn over = Σf_ix_i = 19140
Thus, required mean value = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{19140}{1000}\) = 19.14

Question 14.
The mean of the following data is 20.5. Find the missing frequency.

x	10	15	20	25	30
f	5	7	•••	12	6

Solution:
Let the missing frequency be f₁
The table of values is given as under :

X	f	fx
10	5	50
15	7	105
20	f1	20f1
25	12	300
30	6	180
	Σf = 30 + f1	Σfx = 635 + 20f1

∴ by direct method, Mean = \(\frac{\Sigma f x}{\Sigma f}\) ⇒ Mean = \(\frac{635+20 f_1}{30+f_1}\)
Also given mean be 20.5.
∴ 20.5 = \(\frac{635+20 f_1}{30+f_1}\)
⇒ 20.5 = (30 + f₁) = 635 + 20f₁
⇒ 615 + 20.5 f₁ = 625 + 20f₁
⇒ 0.5 f₁ = 20 ⇒ f₁ = 40
Hence the required missing frequency be 40.

Question 15.
The mean of 40 observations was 160. It was detected on re-checking that the value 125 was wrongly copied as 165 for the computation of the mean. Find the
correct mean.
Solution:
Given no. of observation = 40

Question 16.
The mean of the following frequency table is 50. But the frequencies f₁ and f₂ in classes 20 – 40 and 60 – 80 are missing. Find the missing frequencies.

Class	0-20	20-40	40-60	60-80	80 -100
Frequency	19	f1	32	f2	19	(Total 120)

Solution:

Class	frequency fi	Mid-class xi	fixi
0-20	19	10	190
20-40	f1	30	30f1
40-60	32	50	1600
60-80	f2	70	70f2
80-100	19	90	1710
Total	70 + f1 + f2 = Σf		Σfixi = 3500 + 30f1 + 70f2

Given Σf_i = 120 ⇒ 120 = 70 + f₁ + f₂ ⇒ f₁ + f₂ = 50 …(1)
By direct method, we have
Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) ⇒ 50 = \(\frac{3500+30 f_1+70 f_2}{120}\) ⇒ 3500 + 30f₁ + 70f₂ = 6000
⇒ 30f₁ + 70f₂ = 2500 ⇒ 3f₁ + 7f₂ = 250 …(2)
On solving eqn. (1) by eqn. (2); we have
4f₂ = 100 ⇒ f₂ = 25 and f₁ = 25

Question 17.
The mean of 30 values was 150. It was detected on re-checking that the value 165 was wrongly copied as 135 for the computation of the mean. Find the correct mean.
Solution:
Given no. of observations = 30

Question 18.
The sum of deviation of set of values x₁, x₂,…, x_n measured from 50 is – 10 and the sum of deviation of values from 46 is 70. Find the value of n and the mean.
Solution:

Students can cross-reference their work with Class 11 ISC Maths S Chand Solutions Chapter 17 Circle Chapter Test to ensure accuracy.

S Chand Class 11 ICSE Maths Solutions Chapter 17 Circle Chapter Test

Question 1.
Find the centre and radius of the circle
2x² + 2y² – x = 0.
Solution:
Given eqn. of circle be
2x² + 2y² – x = 0 ⇒ x² + y² – \(\frac { x }{ 2 }\) = 0
∴ its centre be \(\left(\frac{1}{4}, 0\right)\)
and radius of circle = r = \(\sqrt{\left(\frac{1}{4}\right)^2+0}\) = \(\frac { 1 }{ 4 }\)

Question 2.
Find the equation of the circle with centre (- a, – b) and radius \(\sqrt{a^2-b^2}\).
Solution:
Using centre-radius form, eqn. of circle having centre (- a, – b) and radius \(\sqrt{a^2-b^2}\) is given by
(x + a)² + (y + b)² = (\(\sqrt{a^2-b^2}\))²
⇒ x² + y² + 2ax + 2by + a² + b² = a² – b²
⇒ x² + y² + 2ax + 2by + 2b² = 0

Question 3.
Find the equation of the circle drawn on the line joining (- 1, 2) and (3, – 4) as diameter.
Sol.
Using diameter form, eqn. of circle having the points (- 1, 2) and (3, – 4) as extremities of diameter is given by
(x + 1) (x – 3) + (y – 2) (y + 4) = 0
⇒ x² + y² – 2x + 2y – 11 = 0
which is the required eqn. of circle.

Question 4.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre lies on the line 4x + y = 16.
Solution:
Let the general eqn. of circle be given by
x² + y² + 2gx + 2fy + c = 0
Now circle (1) passes through the points (4, 1) and (6, 5).
16 + 1 + 8g + 2f + c = 0
⇒ 8g + 2f + c + 17 = 0 …(2)
36 + 25 + 12g + 10f + c = 0
⇒ 12g + 10f + c + 61 = 0
The centre (- g, – f) of circle (1) lies on line 4x + y = 16
∴ – 4g – f = 16
⇒ 4g + f + 16 = 0 …(3)
eqn. (3) – eqn. (2) gives;
4g + 8f + 44 = 0
⇒ g + 2f + 11 = 0
On solving eqn. (4) and eqn. (5); we have
7g + 21 = 0 ⇒ g = – 3 and f = – 4
∴ from (2) ;
– 24 – 8 + c + 17 = 0 ⇒ c = 15
putting the values of g, f and c in eqn. (1); we have
x² + y² – 6x – 8y + 15 = 0
which is the required eqn. of circle.

Question 5.
Find the equation of a circle of radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0
Since centre of (1) i.e. (- g, – f) lies on x-axis.
∴ – f = 0 ⇒ f = 0
∴ eqn. (1) reduces to;
x² + y² + 2gx + c = 0 …(2)
Now circle given by eqn. (2) passes through the point (2, 3).
4 + 9 + 4g + c = 0
⇒ 4g + c + 13 = 0 …(3)
and radius of circle (2) = \(\sqrt{g²-c}\)
Also, given radius of circle be 5 .
∴ g² – c = 25 …(4)
On adding eqn. (3) and (4); we have
4g + g² – 12 = 0
⇒ (g – 2)(g + 6) = 0 ⇒ g = 2, – 6
When g = 2 ∴ from (3) c = – 21
Thus eqn. (2) reduces to;
x² + y² + 4x – 21 = 0
When g = – 6 ∴ from (3); c = 11
∴ eqn. (2) reduces to ;
x² + y² – 12x + 11 = 0
Thus eqns. (5) and (6) gives the required eqns. of circles.

Question 6.
Find the equation of the circle concentric with the circle x² + y² – 8x + 6y – 5 = 0 and passing through the point (- 2, – 7).
Solution:
eqn. of given circle be
x² + y² – 8x + 6y – 5 = 0
∴ Centre of circle (1) be (4, – 3)
and radius of circle = \(\sqrt{16+9+5}\) = \(\sqrt{30}\)
Since the required circle is concentric with circle (1).
∴ Centre of required circle be (4, – 3) and required circle passed through the point (- 2, – 7)
∴ radius of required circle
= \(\sqrt{(-2-4)^2+(-7+3)^2}\)
= \(\sqrt{36+16}\) = \(\sqrt{52}\)
Hence the required eqn. of circle having centre (4, – 3) and radius \(\sqrt{52}\) be given by
(x – 4)² + (y + 3)² = 52
⇒ x² + y² – 8x + 6y – 27 = 0

Question 7.
Find the equation of the circle through the points (0, 0), (2, 0) and (0, 4). Also find the coordinates of its centre and its radius.
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0
Circle (1) passes through the point (0, 0).
∴ 0 + 0 + 0 + 0 + c = 0 ⇒ c = 0
Further the points (2, 0) and (0, 4) also lies on eqn. (1).
∴ 4 + 0 + 4g = 0 ⇒ g = – 1
and 16 + 0 + 8f = 0 ⇒ f = – 2
putting the values of g, f and c in eqn. (1) ; we get
x² + y² – 2x – 4y = 0
which is the required eqn. of circle.
Further centre of circle be (- g, – f) i.e. (1, 2)
and radius of circle = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{1+4-0}\) = \(\sqrt{5}\)

Question 8.
Find the parametric representation of the circle x² + y² – 2x – 4y – 4 = 0.
Solution:
Given eqn. of circle be
x² + y² – 2x – 4y – 4 = 0
⇒ x² – 2x + y² – 4y – 4 = 0
⇒ (x² – 2x + 1) + (y² – 4y + 4) = 9
⇒ (x – 1)² + (y – 2)² = 9
⇒ \(\left(\frac{x-1}{3}\right)^2\) + \(\left(\frac{y-2}{3}\right)^2\) = 1 …(1)
Thus its parametric eqns. are ;
\(\frac { x – 1 }{ 3 }\) = cos θ ⇒ x = 1 + 3 cos θ
and \(\frac { y – 2 }{ 3 }\) = sin θ ⇒ y = 2 + 3 sin θ where θ be the parameter.

Question 9.
Find the length of the chord intercepted by the circle x² + y² = 25 on the line 2x – y + 5 = 0.
Solution:
Given eqn. of circle be x² + y² = 25 its centre be C(0, 0) and radius = 5
From C, drawn CM ⊥ chord AB Clearly it bisects the chord.

In right angled △AMC, we have
AC² = AM² + CM²
⇒ AM² = 5² – CM²
| CM | = length of ⊥ drawn from C(0, 0) to line 2x – y + 5 = 0
= \(\frac{|2 \times 0-0+5|}{\sqrt{2^2+(-1)^2}}\) = \(\frac{5}{\sqrt{5}}\) = \(\sqrt{5}\)
∴ from (1); AM² = 25 – 5
⇒ AM = \(\sqrt{20}\) = \(2 \sqrt{5}\)
∴ required length of chord intercepted by given circle on given line = AB = 2AM = 4\(\sqrt{5}\) units

Question 10.
Find the equations of the tangents to the circle x² + y² = 9, which are parallel to the line 3x + 4y = 0.
Solution:
Given eqn. of circle be
x² + y² = 9 …(1)
Its centre be C(0, 0) and radii 3 units and eqn. of given line be
3x + 4y = 0 …(2)
∴ eqn. of line parallel to line (2) be given by
3x + 4y + k = 0 …(3)
Since line (3) is tangent to circle (1).
∴ ⊥ distance drawn from C(0, 0) of circle to line (3) = radius of circle.
⇒ \(\frac{|3 \times 0+4 \times 0+k|}{\sqrt{3^2+4^2}}\) = 3 ⇒ \(\frac{|k|}{5}\) = 3
⇒ k = ± 15
∴ from (3) ; 3x + 4y ± 15 = 0
be the required equations of tangents to given circle.

Question 11.
Find the equation of the circle which touches the y-axis at a distance of +4 from the origin and cuts off an intercept 6 from the x-axis.
Solution:
There are two such circles one in first and other in second quadrant satisfying the given conditions.
Let C and D be the centres of both circles.
From C draw CM ⊥ PQ
∴ M be the mid-point of PQ.
MP = MQ = 3 units since PQ = 6 units

∴ In right angled △CPM, we have
CP² = CM² + PM² = 4² + 3² = 25
⇒ CP = 5 = radius of circle
⇒ CB = 5 units ⇒ OM = 5 units
∴ the coordinates of C and D are (5, 4) and (- 5, 4)
Thus required eqns. of circles are given by
(x ± 5)² + (y – 4)² = 5²
⇒ x² + y² ± 10 x – 8y + 16 = 0