Oswal 61 Sample Question Papers ICSE Class 9 Maths Solutions
Section-A
Answer 1.
(i) (c) rational number
(π – π) = 0 is a rational number.
$$x^6 – 27x^3 + 26 = (x^3)^2 – 27x^3 + 26\\\text{Let}\space x^3=y\\=y^2 – 27y + 26\\=y^2 – 26y – y + 26\\= y(y – 26) – 1(y – 26)\\= (y – 26)(y – 1)\\= (x^3 – 26)(x^3 – 1) [\text{Put the value of y} = x^3]\\= (x3 – 26)[x3 – (1)^3]\\=(x^3-26)(x-1)(x^2+x+1)\\\space[\because a^3-b^3=(a-b)(a^2+b^2+ab)]\\=(x – 1)(x^2 + x + 1)(x^3 – 26)$$
(iv) (d) 0
2p = 3q = 6r = k 2p = k
$$\Rarr 2=k^{1/p}\space…(i)\\3^q=k\\\Rarr 3=k^{1/q}\space…(ii)\\6^r=k\\\Rarr 6=k^{1/r}\space…(iii)\\\therefore 2×3=k^{1/r}\\\Rarr k^{1/p}.k^{1/q}=k^{1/r}\space((\text{from equation (i) and (ii)}))\\\Rarr K^{{1/p}+{1/q}}=k^{1/r}\\\Rarr\frac{1}{p}+\frac{1}{q}=\frac{1}{r}\\\Rarr\frac{1}{p}+\frac{1}{q}-\frac{1}{r}=0$$
(v) (a) 2 log a
3 log a – log a = 2 log a
(vi) (b) PQ > QR

$$\because PR=QR\\\therefore\angle P=\angle Q\\\because\text{Sum of all interior angles of triangle = 180}\degree \therefore \angle P+\angle Q+ \angle R=180\degree\\\Rarr\space 37\degree+37\degree+\angle R=180\degree\\\Rarr\angle R=180\degree-74\degree\\\Rarr\angle R=106\degree\\\because\text{In a triangle angle opposite to greater side is greater and vice-versa.}\\\therefore\angle R\space\text{\textgreater}\angle P\\\Rarr PQ\text{\textgreater}QR$$
(vii) (b) parallelogram
By mid-point theorem,

$$\text{PQ}||\text{AC}\\\qquad\text{PQ}=\frac{1}{2}\text{AC}\space…(i)\\\text{RS}||\text{AC}\\\text{RS}=\frac{1}{2}\text{AC}\space…\text{(ii)}\\\text{From equation (i) \text{and}\space \text{equation}\space (ii),\text{we get}}\\\text{PQ=RS}\\\text{PQ||RS}\\\therefore\text{One pair of opposite side are equal and parallel.}\\\text{Hence, PQRS is a parallelogram.}$$
(viii) (c) 40°
In the adjoining figure,
∠OAD = ∠OCB (Alternate angles)
∠OCB = 30°
∠AOB + ∠BOC = 180°
⇒ 70° + ∠BOC = 180°
⇒ ∠BOC = 180° – 70°
⇒ ∠BOC = 110°
In ΔBOC
∠OBC + ∠BOC + ∠OCB = 180°
⇒ ∠OBC + 110° + 30° = 180°
⇒ ∠OBC = 40°
⇒ ∠DBC = 40°
(ix) (b) 1 : 1

We know,
Parallelograms on the same base and between the same parallel lines are equal in area (theorem).
Area of □ABCD = Area of □ABEF
Ratio = 1 : 1
(x) (b) False
Given, AD is diameter. AB can never be equal to AD because diameter is the largest chord of circle.
Hence AD > AB.

Also, to make AB = AD, AD should pass through centre of circle.
(xi) (b) 16
Mean of 45 numbers = 18
⇒ Sum of 45 numbers = 18 × 45 = 810
Mean of (75 – 45) numbers = 13
Sum of 30 numbers = 13 × 30 = 390
Sum of 75 numbers = Sum of 45 numbers + Sum of 30 numbers.
Sum of 75 numbers = 810 + 390 = 1200
X=1200/75
=16
BC = 19 cm

(xiii) (b) 70°
3 cot2 (x – 10°) = 1
⇒ cot2(x – 10°) = 1/3
⇒ cot(x – 10°) = 1/√3
We know, cot 60° = 1/√3
⇒ cot(x – 10°) = cot 60°
⇒ x – 10° = 60°
⇒ x = 70°
(xiv) (c) 10 √3 units
Draw perpendiculars on AB from DC.

In DDXA,
sin 60° = DX /DA
⇒ √3/2 = DX /20
⇒ DX = 10√3
⇒ Distance between AB and DC is 10√3 uunniittss.
(xv) (c) Scalene triangle
$$\text{Let,}\space A(8,3),B(0,9),C(14,10)\\\text{AB}=\sqrt{(0-8)^2+(9-3)^2}\\=\sqrt{64+36}=10\\\text{BC}=\sqrt{(14-0)^2+(10-9)^2}\\=\sqrt{196+1}=\sqrt{197}\\AC=\sqrt{(14-8)^2+(10-3)^2}\\=\sqrt{36+49}=\sqrt{85}\\\text{Since},AB\neq BC\neq AC.\\\Rarr\text{A, B, C are coordinates of scalene triangle.}$$
Answer 2.
- (i) We know that,
- (a + b)3 = a3 + b3 + 3a2b + 3ab2
- and (a – b)3 = a3 – b3 – 3a2b + 3ab2
- (a + b)3 – (a – b)3 = 2[b3 + 3a2b]
- (x/2+ y/3)3-(x/2 - y/3)3=2[(y/3)3+3[(y/3)3(y/3)]
- =2(y3/27+x2y/4)
- =2y3/27+x2y/2
- (ii) 5 – (3a2 – 2a) (6 – 3a2 + 2a) = 5 – (3a2 – 2a) [6 – (3a2 – 2a)]
- Let, 3a2 – 2a = x
- 5 – (3a2 – 2a) (6– 3a2 + 2a) = 5– x (6 – x) = 5 – 6x + x2
- = 5 – 5x – x + x2
- = 5(1 – x) – x (1 – x)
- = (5 – x) (1 – x)
- = (x – 5) (x – 1)
- = (3a2 – 2a – 5) (3a2 – 2a – 1) [Put x = 3a2 – 2a]
- = (3a2 – 5a + 3a – 5) (3a2 – 3a + a – 1)
- = {a (3a – 5) + 1 (3a – 5)} {3a (a – 1) + 1 (a – 1)}
- = (3a – 5) (a + 1) (3a + 1) (a – 1)
- (iii) Let the larger number be x and the smaller number be y.
- We know that,
- Dividend = Divisor × Quotient + Remainder
- 3x = (y × 4) + 3
- ⇒ 3x – 4y = 3 ...(i)
- and 7y = (x × 5) + 1
- ⇒ 5x – 7y = – 1 ...(ii)
- From equation (i), we have
- 3x = 4y + 3
- ⇒ x =4y=3/3 ...(iii)
- Substituting the value of x in equation (ii), we get
- 5((4y+3)/3)-7y=-1
- ⇒ 20y + 15 – 21y = – 3
- ⇒ – y = – 18
- ⇒ y = 18
- Putting the value of y in equation (iii), we get
- x =(4 x 18 + 3)/3=(72+3)/3=(75)/3=25
- x = 25 and y = 18
- The two numbers are 25 and 18. Ans.
Answer 3.
- A = P(1+(r/100 x 2))2n
- ⇒ 4,576 = 4,400 (1+(8/100 x 2))2n
- ⇒ 4,576 = 4,400()26/25)2n
- ⇒ (26/25)2n=4,576/4,400
- ⇒ (26/25)2n=26/25
- ⇒ (1·04)2n = 1·04
- ⇒ (1·04)2n = (1·04)1
- From equation (i), we have
- ⇒ 2n = 1
- ⇒ n =1/2
- Thus, the required time period is half year. Ans.
- (ii) Given : AB = AC, BC = CD and DE || BC.
- (a) In ΔABC,
- AB = AC
- ⇒ ∠ACB = ∠ABC ...(i)
- Also, ∠ABC + ∠ACB = ∠FAB [Exterior angle property]
- ⇒ 2∠ABC = 128° [from equation (i)]
- ⇒ ∠ABC = 128@/2=64°
- ⇒ ∠ABC = ∠ACB = 64°
- InΔBCD,
- BC = CD [Given]
- ⇒ ∠BDC = ∠DBC ...(ii)
- Also, ∠BDC + ∠DBC + ∠BCD = 180° [Angle sum property]
- ⇒ 64° + 64° + ∠BCD = 180°
- ∠BCD = 180° – 128° = 52°
- Now, DE || BC
- ∠CDE = 52° Ans.
- (b) ∠ACB = ∠BCD + ∠DCE
- ⇒ ∠DCE = ∠ACB – ∠BCD
- = 64° – 52°
- ⇒ ∠DCE = 12°. Ans.
- (iii) We have,
- 5x – (5 – x) = 1/2 (3− x)
- ⇒ 5x – 5 + x = 1/2 (3− x)
- ⇒ 2(6x – 5) = 3 – x
- ⇒ 12x – 10 = 3 – x
- ⇒ 13x = 13
- x = 13 /13 =1
- Ans, 4 – 3y =(4y+y)/3
- ⇒ 3(4 – 3y) = 4 + y
- ⇒ 12 – 9y = 4 + y
- ⇒ – 9y – y = 4 – 12
- ⇒ – 10y = – 8
- ⇒ y =-18/-10=4/5
- So, the coordinates of the point are (1,4/5) Ans.
Section-B
Answer 4.
(i) Let us draw a right triangle ABC, right-angled at C in which tan A =1/√3

- Now, tan A = BC/AC=1/√3
- Let BC = x
- and AC = √3 x
- By Pythagoras theorem, we
- AB2= AC2 + BC2
- ⇒ AB2 = (√ 3x)2 + x2
- ⇒ AB2 = 3x2 + x2
- ⇒ AB2 = 4x2
- ∴ AB = 2x
- With reference to ∠A, we have
- Base AC =√3x, Perpendicular BC=x and Hypotenuse AB=2x
- sin A = BC/AB = x/2x=1/2
- and, cosA =AC/AB
- =√3x/2x=√3/2
- With reference to ∠B, we have
- Base BC = x, Perpendicular AC = √3x and Hypotenuse AB = 2x.
- ∴ cos B =BC/AB=x/2x=1/2
- and sin B =AC/AB=√3x/2x=√3/2
- Now, sin A. cos B + cos A. sin B
- =1/2 x 1/2 + √3/2 x √3/2
- =1/4+3/4=1 Ans.
- (ii) ·.· Length of rectangular room = x
- ∴ Width = 4x/7
- ⇒ 2(x+4x/7)=y
- ⇒ 2(11x/7)=y
- ⇒ (22x/7)=y Ans.
- ∴ 22 x/7=4400
- ⇒ x = 1400
- Length of the room is 1400 cm or 14 m. Ans.
- (iii) (a) Class mark = ( Lower limit +Upper limit )/2
- =(90+120)/2=210/2=105 Ans.
- (b) Lower limit = Mid-value – Width of class/2 = 10-6/2
- = 10 – 3
- Lower limit of class = 7 Ans.
- (c) Given : Lower limit of the lowest class = 10
- Also, Width of each class = 5
- and Total number of classes = 5
- 1st class = 10 – 15, 2nd class = 15 – 20
- 3rd class = 20 – 25, 4th class = 25 – 30
- and 5th class = 30 – 35
- Upper limit of highest class = 35 Ans.
Answer 5.
(i) Given : In DABC, AD is the median.
To prove : AB + AC > 2 AD

- Construction : Produce AD upto a point E such that AD = DE and join C to E.
- Proof : In ΔACE,
- AC + CE > AE
- [... sum of two sides of a triangle is greater than the third side]
- ⇒ AC + CE > 2 AD [ AD = DE] ...(i)
- In ΔADB and ΔCDE,
- BD = CD [Given]
- AD = DE [By construction]
- ∠ADB = ∠CDE [Vertically opposite angle]
- ∠ADB ≅ ∠CDE [By SAS
- ∠ADB ≅ ∠CDE [By SAS congruencey rule]
- So, AB = CE [C.P.C.T.]So, AB = CE [C.P.C.T.]
- Now, substituting CE = AB in equation (i), we have
- AC + AB > 2 AD
- ⇒ AB + AC > 2 AD Hence Proved.
- (ii) Let us consider that √5 is a rational number.
- Then, let √5 =p/q, .....(i)
- where p and q are co-prime integers such that q ≠ 0, p and q have no common factor, except 1. Squaring on both sides in equation (i), we get
- ∴ 5 =p2/q2
- ⇒ p2 = 5q2 …(ii)
- ·.· 5q2 is divisible by 5.
- ∴ p2 is also divisible by 5.
- ∴ p is also divisible by 5.
- Let p = 5k, where k is an integer.
- ∴ p2 = 25k2
- ⇒ 5q2 = 25k2 [By using equation (ii)]
- ⇒ q2 = 5k2
- ·.· 5k2 is divisible by 5.
- ∴ q2 is also divisible by 5.
- ∴ q is also divisible by 5.
- ⇒ p and q both are divisible by 5.
- ⇒ p and q both multiples of 5.
- But p and q must have no common factor other than 1.
- It means our assumption is wrong.
- ∴ √5 is an irrational number. Hence Proved.
- (iii) Given : C is the mid-point of AB, ∠DCA = ∠ECB and ∠DBC = ∠EAC.
- To prove : DC = EC
- Proof : We have, ∠DCA = ∠ECB [Given]
- We have, ∠DCA + ∠DCE = ∠ECB + ∠DCE [Adding ∠DCE to both sides]
- ∠ACE = ∠DCB
- Now, in DACE and DDCB
- ∠ACE = ∠DCB [Proved above]
- AC = CB [Given]
- ∠CAE = ∠CBD [Given]
- By ASA congruency rule,
- ∠ACE ≅ DBCD
- DC = EC Hence Proved.
Answer 6.
$$\text{(i) Cost of the T.V. in 2001= P}\bigg(1+\frac{R_1}{100}\bigg)\bigg(1-\frac{R_2}{100}\bigg)\\=17,000\bigg(1+\frac{5}{100}\bigg)\bigg(1-\frac{4}{100}\bigg)\\=17,000 ×\frac{21}{20}×\frac{24}{25}\\= 17,136\\\text{Thus, the cost of the T.V. set in 2001 was}\space ₹ 17,136. \textbf{Ans.}$$
- (ii) Given : PQ = PX,
- PR = PS
- and ∠QPX = ∠RPS
- To prove : QR = XS
- Proof : ∠QPX = ∠RPS
- ⇒ ∠QPX + ∠XPR = ∠RPS + ∠XPR [On adding ∠XPR both sides]
- ⇒ ∠QPR = ∠XPS ...(i)
- In DQPR and DXPS,
- PQ = PX [Given]
- ∠QPR = ∠XPS and PR = PS [Given]
- By SAS congruency rule
- DQPR ≅ DXPS
- QR = XS [C.P.C.T.] Hence Proved.
Answer 7.
$$\text{(i) First ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29}\\\text{Mean}=\frac{2+3+5+7+11+13+17+19+23+29}{10}\\=\frac{129}{10}=12.9\\\text{Since, number of term (n) is even,}\\\therefore\text{Median}\frac{1}{2}\bigg[\bigg(\frac{n}{2}\bigg)^{th}\text{term}+\bigg(\frac{n}{2}+1\bigg)^{th}\text{term}\bigg]=\frac{1}{2}[5\text{th term} + 6\text{th term}]=\frac{1}{2}[11+13]=12\\\therefore\text{Mean}=12.9\space\text{and median = 12}\textbf{Ans.}$$
$$\text{(ii) Given, A + B = 90}\degree\\\Rarr B=90\degree-A\space\text{…(i)}\\\text{Now, L.H.S.}\\=\sqrt{\frac{\text{tan A tan B} +\text{tan A cot B}}{\text{sin AsecB}}-\frac{\text{sin}^{2}B}{\text{cos}^{2}A}}\\=\sqrt{\frac{\text{tan A tan}(90\degree-A)+\text{tanA cot}(90\degree-A)}{\text{sin Asec}(90\degree-A)}-\frac{\text{sin}^{2}(90\degree-A)}{\text{cos}^{2}A}}\\=\sqrt{\frac{\text{tanAcotA}+\text{tanAtanA}}{\text{sin AcosecA}}-\frac{\text{cos}^{2}A}{\text{cos}^{2}A}}\\\begin{bmatrix}\therefore\space\text{sin}(90\degree-\theta)=\text{cos}\theta\\\text{cot}(90\degree-\theta)=\text{tan}\theta\\\text{sec}(90\degree-\theta)=\text{cosec}\theta\end{bmatrix}\\=\sqrt{\frac{\text{tan} A×\frac{1}{tan A}+\text{tan}^{2}A}{\text{sinA}×\frac{1}{sin A}}}-1\space\space\space\space\begin{bmatrix}\because\space\text{cos}\theta =\frac{1}{tan \theta}\\\text{cosec}\theta=\frac{1}{sin \theta}\end{bmatrix}\\=\sqrt{1+\text{tan}^{2}A-1}\\=\sqrt{\text{tan}^{2}A}\\=\text{tanA =R.H.S}\space\textbf{Hence Proved.}\\\text{(iii) In ∆ABD, using Pythagoras theorem,}\\\text{AB}=\sqrt{AD^{2}-BD^{2}}\\\text{AB}=\sqrt{26^{2}-24^{2}}\\\sqrt{100}=10cm\\\text{The area of right triangle ABD will be}\\\therefore\text{ar}(\Delta ABD) =\frac{1}{2}×AB×BD\\=\frac{1}{2}×10×24=120\text{cm}^{2}\\\text{Also}\space\text{ar}(\Delta BCD)=\frac{\sqrt{3}}{4}×(\text{Side})^2\\=\frac{\sqrt{3}}{4}×(24)^2\\= 144\sqrt{3}\space cm^2\\\text{Now, Area of quad. ABCD} = ar(\Delta ABD) + ar (\Delta BCD)\\= 120 + 144\sqrt{3}\\= 120 + 249·41\\= 369.41 \text{cm}^{2}\space\textbf{Ans.}$$
$$=\frac{1}{2}×10×24=120\text{cm}^{2}\\\text{Also}\space\text{ar}(\Delta BCD)=\frac{\sqrt{3}}{4}×(\text{Side})^2\\=\frac{\sqrt{3}}{4}×(24)^2\\= 144\sqrt{3}\space cm^2\\\text{Now, Area of quad. ABCD} = ar(\Delta ABD) + ar (\Delta BCD)\\= 120 + 144\sqrt{3}\\= 120 + 249·41\\= 369.41 \text{cm}^{2}\space\textbf{Ans.}$$
Answer 8.

$$\text{(i)\space We have,}\text{cosec} A=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{\sqrt{10}}{1}\\\text{So, we draw a right triangle ABC, right-angled at B, such that Perpendicular BC = 1 unit and Hypotenuse AC =}\sqrt{10}\space\text{units.}\\\text{By Pythagoras theorem, we have}\\\text{AC}^{2} = AB^{2} + BC^{2}\\\Rarr(\sqrt{10})^{2}=AB^{2}+1^{2}\\\Rarr\space AB^{2} = 10 – 1 = 9\\\therefore\text{AB}=\sqrt{9}=3\\\text{With reference to}\space\angle A,\text{we have}\\\text{Base\space AB=3},\text{Perpendicular BC = 1 and Hypotenuse}\\\text{AC}=\sqrt{10}\\\therefore\text{sin A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{1}{\sqrt{10}}\\$$
$$\text{cos A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{3}{\sqrt{10}}\\\text{tan}A=\frac{\text{Perpendicular}}{\text{Base}}=\frac{1}{3}\\\qquad\text{sec A}=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\sqrt{10}}{3}\\\text{and}\space\text{cot} A=\frac{Base}{\text{Perpendicular}}=\frac{3}{1}\\=3\space \textbf{Ans.}\\\text{(ii)}\space\bigg(\frac{1}{4}\bigg)^{-2}-3(8)^{2/3}×4^{0}+\bigg(\frac{9}{16}\bigg)^{-1/2}=(4)^{2}-3×2^{3×2/3}×1+\bigg(\frac{4}{3}\bigg)^{2×1/2}\\=16- 12+\frac{4}{3}=4+1\frac{1}{3}=5\frac{1}{3}\space\textbf{Ans.}\\$$

$$\text{(iii) Let AB and CD be the two poles such that} AB \text{\textless} CD.\\\Rarr\space AB = 6 m, CD = 11 m\\\text{and}\space AC=12m\\\textbf{Construction} : \text{Draw}\\BP\perp CD\\\text{Now,}\text{in}\space\angle BPD,\\\angle P=90\degree\\\Rarr BD^{2}=BP^{2}+PD^{2}\space [\text{Pythagoras theorem}]\\(BD^{2})=(AC^{2})+(CD-CP)^2\space[\because BP=AC]\\\Rarr BD^{2}=(12)^{2}+(11-6)^2\\\Rarr BD^2=144+25\\\Rarr BD=\sqrt{169}=13m\\\text{Hence, the distance between their tops is 13 m.} \textbf{Ans.}$$
Answer 9.
$$\text{(i) (a) Area of}\Delta\space ABC=\frac{1}{2}×BC×AB=\frac{1}{2}×12×16=96\space\text{cm}^{2}\space\textbf{Ans.}\\\text{(b) In} \Delta ACD,\text{semi perimeter} s=\frac{a+b+c}{2}\\=\frac{20+29+21}{2}=\frac{70}{2}=25\\\therefore\text{Area of}\space\Delta ACD=\sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{35(35-20)(35-29)(35-21)}\\=\sqrt{35×15×6×14}\\=\sqrt{7×5×3×5×2×3×17×2}\\=2×3×5×7=210\space\text{cm}^{2}\textbf{Ans.}\\\text{(c)\space Area of quadrilateral = Area of}\space\Delta \text{ABC}+\text{Area of}\space\Delta\text{ACD}\\=96+210=306\text{cm}^{2}\textbf{Ans.}$$
- (ii) The given frequency distribution is in inclusive form, so first we convert it into exclusive form, as shown below.
Height (in cm) | Number of children (Frequency) |
120.5—130.5 | 10 |
130.5—140.5 | 15 |
140.5—150.5 | 12 |
150.5—160.5 | 16 |
160.5—170.5 | 8 |
170.5—180.5 | 4 |

Answer 10.
- (i) Given, S.I. for 2 years = ₹ 200
- ⇒ P × 10 × 2/100=200
- ⇒ P = ₹ 1,000
- Now, C.I. for the first year
- = 1, 000 x 10 x 1/100=₹100
- ∴ Amount at the end of first year = `₹1,000 + ₹ 100 = ₹1,100
- C.I. for the second year =1,100 x 10 x1/100= ₹110
- ∴ Amount of the end of second year = ₹ 1,100 + ₹ 110 = ₹1,210
- So, Total C.I. in 2 years = ₹ 1,210 – ₹ 1,000 = ₹ 210. Ans.
- (ii) Given : ∠PRQ = 75°, ∠PSQ = 36° and PQ = PR
- ∠PRQ = ∠PQR = 75° [... PQ = QR]
- In ΔPQR,
- ∠PQR + ∠PRQ + ∠QPR = 180° [Angle sum property]
- ⇒ 75° + 75° + ∠QPR = 180°
- ⇒ ∠QPR = 180° – 150° = 30°
- ∠QPR < ∠PQR
- QR < PR ...(i)
- [ Side opposite to smaller angle is shorter]
- ∠RPS + ∠PSR = ∠PRQ [Exterior angle property]
- ⇒ ∠RPS + 36° = 75°
- ⇒ ∠RPS = 75° – 36° = 39°
- ∠PSR < ∠RPS
- PR < RS ...(ii)
- From equation (i) and equation (ii), we get
- QR < PR < RS Ans.
- (iii) Let O (x, y) be the centre of the circle.
- Also, let A (6, – 6), B (3, – 7), C (3, 3) be the three points which lie on the circle.

- ⇒ (6 – x)2 + (– 6 – y)2 = (3 – x)2 + (– 7 – y)2
- ⇒ 36 + x2 – 12x + 36 + y2 + 12y = 9 + x2 – 6x + 49 + y2 + 14y
- ⇒ – 12x + 6x + 12y – 14y + 72 – 58 = 0
- ⇒ – 6x – 2y + 14 = 0
- ⇒ – 2 (3x + y – 7) = 0
- ⇒ 3x + y = 7 ...(i)
- Similarly, OA = OC
- or OA2 = OC2
- ⇒ (6 – x )2 + (– 6 – y)2 = (3 – x)2 + (3 – y)2
- ⇒ 36 + x2 – 12x + 36 + y2 + 12y = 9 + x2 – 6x + 9 + y2 – 6y
- ⇒ – 12x + 6x + 12y + 6y + 72 – 18 = 0
- ⇒ – 6x + 18y + 54 = 0
- ⇒ – 6 (x – 3y – 9) = 0
- ⇒ x – 3y = 9 ...(ii)
- Solving equations (i) and (ii), we get
- x = 3 y = – 2
- Hence, the centre of circle is (3, – 2). Ans.
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