NCERT Solutions of class 11 maths exercise 11.1 chapter 11-Conic sections - Future Study Point

# NCERT Solutions of class 11 maths exercise 11.1-Conic sections

NCERT Solutions of class 11 maths chapter 11 -Conic sections are the solutions of unsolved questions of the NCERT maths text book of class 11 chapter 11-Conic sections  prescribed by  CBSE a reputed school board of India.All NCERT Solutions of class 11 maths chapter 11 are solved by CBSE maths expert by a step by step way. All NCERT solutions of class 11 maths chapter 11 are explained here will help all maths students of class 11 and the candidates who are going to appear in competitive entrance exams like NDA,CDS,engineering entrance exams etc .

PDF-NCERT solutions class 11 chapter 11-Conic Section

The chapter 11 of NCERT maths text book of class 11 is based on the –Conic Section, Conic Sections contains parabola,ellipse and hyperbola which are the planes carved out from the cone,it is that’s why chapter is named ‘Conic Section’. The questions are based on the equations of all the figure of parabola,ellipse and hyperbola.The chapter ‘Conic Section’ is easy so students can get an excellent score in maths by attempting the questions of chapter 11-Conic Sections in the maths question paper of class 11.

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Q1.FNCERT Solutions of class 11 maths chapter 11ind the equation of the circle with center (0,2) and radius 2.

Solution. The standard equation of the circle  is

(x – h)² + (y – k)² = r²

Where (h,k) are the coordinates of the circle and r is the radius of the circle

It is given that center (h,k) = (0,2) and radius (r)=2

Therefore, equation of the circle is

(x – 0)² + (y – 2)² = (2)²

x² + y² + 4 – 4y = 4

x² + y² – 4y = 0

Q2.Find the equation of the circle with center (-2,3) and radius 4.

Ans. The equation of the circle with centre (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (-2,3) and radius (r)= 4

Therefore, equation of the circle is

(x + 2)² + (y – 3)² = (4)²

x² + 4x² + 4 + y² – 6y + 9 = 16

x² + y² + 4x – 6y – 3 = 0

Q3. Find the equation of the circle with the centre (1/2, 1/4) and radius 1/12.

Ans. The equation of the circle with center (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (-1/2,1/4) and radius (r)= 1/12

Therefore, the equation of the circle is

144x² – 144x+ 36 + 144y² – 72y +9 – 1 = 0

144x² – 144x + 144y² – 72y + 44 = 0

36x² – 36x+ 36y² – 18y + 11 = 0

36x² + 36y² – 36x – 18y + 11 = 0

Q4. Find the equation of the circle with the center (1,1) and radius √2.

Ans. The equation of the circle with center (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (1,1) and radius (r)=√2

Therefore, the equation of the circle is

(x – 1)² + (y – 1)² = (√2)²

x² – 2x +1 + y² – 2y + 1 = 2

x² + y² – 2x – 2y = 0

Q5. Find the equation of the circle with the center (-a,-b) and radius √(a² – b²).

Ans. The equation of the circle with center (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (-a,-b) and radius (r)= √(a² – b²)

Therefore, the equation of the circle is

(x + a)² + (y + b)² = [√(a² – b²)]²

x² + 2ax + a² + y² + 2by + b²  =  a² – b²

x² + a² + 2ax + y² + b² + 2by  =  a² – b²

x² + y² + 2ax + 2by + 2b² = 0

Q6. Find the circle and radius of the circle (x + 5)² + (y – 3)² = 36

Ans. The equation of the given circle is

(x + 5)² + (y – 3)² = 36

{x -(- 5)}² + (y – 3)² = 6²

Which is of the form (x – h)² + (y – k)² = r², where h = -5, k = 3, and r = 3.

Thus, the center of given circle is (-5,3) while its radius is 6.

Q7. Find the circle and radius of the circle + y² – 4x – 8y – 45 = 0

Ans. The equation of the circle is

x² + y² – 4x – 8y – 45 = 0

(x² – 4x)+ (y² – 8y) = 45

{x² – 2(x)(2) + 2²} + {y² – 2(y)(4) + 4²} – 4 – 16 = 45

(x – 2)² + (y – 4)² = 65

(x – 2)² + (y – 4)² = (√65)²

Which is of the form (x – h)² + (y – k)² = r², where h = 2, k = 4, and r =√65

Thus, the center of the given circle is (2,4), while its radius is √65.

Q8. Find the center and radius of the circle is x² + y² – 8x +10y – 12 = 0

Ans. Standard equation of the circle is

(x – h)² + (y – k)² = r², where (h,k) are the co-ordinates of the centre of the circle.

The given equation of the circle is x² + y² – 8x +10y – 12 = 0

Rewritting the equation of the circle by applying the complete square method

x²  – 8x + 4²- 4² + y²+10y +5²-5² – 12 = 0

(x – 4)² + (y + 5)² -16 – 25- 12 = 0

(x – 4)² + (y + 5)² = (√53)²

Comparing the standard equation of the circle

(x – h)² + (y – k)² = r²

h =4, k = -5 and r = √53

Therefore, the centre of the circle is (4,-5) and radius of the circle is √53

Q9. Find the centre and radius of the circle 2x² + 2y² -x = 0

Ans. The given equation of the circle is

2x² + 2y² -x = 0

Rearranging the equation by the method of complete square method

2x²  -x + 2y² = 0

Dividing the equation by 2

x² – x/2 + y² = 0

Adding  and subtracting the equation by 1/4

x² -x/2 + 1/4 – 1/4 + y² = 0

(x – 1/2)² + (y – 0)² = (1/2)²

(x – 1/2)² + (y – 0)² = (0.5)²

Now, comparing the equation with standard equation of the circle

(x – h)² + (y – k)² = r², where (h,k) are the co-ordinates of the centre of the circle.

Therefore co-ordinates of the centre are (1/2,0) and radius (0.5)

Q10.Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16

Ans. Let the equation of the required circle be (x – h)² + (y – k)² = r²

Since the circle passes through (4,1) and (6,5)

(4 – h)² + (1 – k)² = r²….(i)

(6 – h)² + (5 – k)² = r²….(ii)

Since the centre (h, k) of the circle lies on line 4x + y = 16

4h + k = 16….(iii)

From equations (i) and (ii), we obtain

(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²

16 – 8h +h² +1 – 2k +k² = 36 – 12h +h² +25 – 10k +k²

16 – 8h +1 – 2k = 36 – 12h +25 – 10k

4h + 8k = 44

h + 2k = 11….(iv)

On solving equations (iii) and (iv), we obtain h = 3 and k = 4

On substituting the values of h and k in equation (i), we obtain

(4 – 3)² + (1 – 4)² = r²

(1)² + (- 3)² = r²

1 + 9 = r²

r² = 10

r = √10

Thus, the equation of the required circle is

(x – 3)² + (y – 4)² = (√10)²

x² – 6x + 9 + y² – 8y + 16 = 0

x² + y² – 6x – 8y + 15 = 0

Q11.Find the equation of the circle passing through the points (2,3) and (-1,1) and whose centre is on the line x – 3y – 11 = 0

Ans. Let the equation of the required circle be (x – h)² + (y – k)² = r²

Since the circle passes thrpugh points (2,3) and (-1,1)

(2 – h)² + (3 – k)² = r²….(i)

(-1 – h)² + (1 – k)² = r²….(ii)

Since the centre (h,k) of the circle lies on line x – 3y – 11 = 0

h – 3k = 11….(iii)

From equations (i) and (ii), we obtain

(2 – h)² + (3 – k)² = (-1 – h)² + (1 – k)²

4 – 4h + h² + 9 – 6k + k² = 1 + 2h + h² + 1 – 2k + k²

4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

6h + 4k = 11….(iv)

On solving equations (iii) and (iv), we obtain h = 7/2 and k = -5/2

On substituting the values of h and k in equation (i), we obtain

(2 – 7/2)² + (3 + 5/2)² = r²

[(4 – 7)/2)]² + [(6 + 5)/2)]² = r²

(-3/2)²  + (11/2)² = r²

9/4 + 121/4 = r²

130/4 = r²

The equation of the required circle is

(x – 7/2)² + (y + 5/2)² = 130/4

[(2x – 7)/2)]² + [(2x + 5)/2)]² = 130/4

4x² – 28x + 49 +4y² + 20y +25 = 130

4x² +4y² – 28x + 20y – 56 = 0

4(x² +y² – 7x + 5y – 14) = 0

x² +y² – 7x + 5y – 14 = 0

∴ The equation of the required circle is x² +y² – 7x + 5y – 14 = 0

Q12.Find the equation of the circle with radius 5 cm whose centre lies on x-axis and passes through the point (2,3)

Ans. The standard equation of the circle is

(x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We know that the radius of the circle is 5 and its centre lies on the x-axis, so coodinates of the centre are (h,k) =(h,0) and r = 5

So now, the equation of the circle is (x – h)² + y ² = 25

It is given that the circle passes through the point (2,3) so the point will satisfy the equation of the circle.

(2 – h)² + 3² = 25

(2 – h)² = 25 – 9

(2 – h)² = 16

2 – h = ± 4

h = 2± 4

h = 6 or -2

The required equation of circle is

(x-6)² +y² =5² or (x + 2)² +k² = 5²

x² + 36 – 12x + y² = 25, x² + 2² +4x +k² = 25

x²  – 12x + y² +11=0, x²  +4x +k² -21 =0

Q13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Ans.The standard equation of the circle is (x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We are given that circle is passing through (0,0) and making intercepts a and b on the coordinate axes means  circle is passing through the point (a,0) on x-axis and (0,b) on y-axis.

substituting (0,0) , (a,0) and (b,0) in the equation of the circle

h² + k² = r²…….(i)

(a -h)² + k²=r²…….(ii)

h² + (b -k)²=r² …….(ii)

from equation (i) and (ii), we have

(a -h)² + k²= h² + k²

a² + h² -2ah + k² = h²  + k²

a² – 2ah = 0 ⇒ a(a – 2h) = 0 ⇒a = 0 and a = 2h

a ≠0, so h = a/2

similarly from equation (iii) and (i), we have

(b -k)² + h²= h² + k²

b² + k² -2bk + h² = h²  + k²

b² – 2bk = 0 ⇒ b(b – 2k) = 0 ⇒b = 0 and b = 2k

b ≠0, so k= b/2

Now, sbstituting the value of (h,k) =(a/2, b/2) in equation (i)

a²/4 + b²/4 = r²

Putting the value of (h,k) and r in the equation of circle

4x²+ 4y² – 4ax – 4by = 0

x²+ y² – ax – by = 0 is the required equation

Q14.Find the equation of a circle with centre (2,2) and passes through point (4,5).

Ans. The standard equation of the circle is (x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We are given that circle is passing through (4,5) and the coordinates(h,k) of the centre are (2,2)

So,we have r² = (4 -2)² + (5 -2)²= 4 + 9 = 13

r = √13

The equation of the circle is

(x -2)² + (y -2)² =13

x ² + 4 -4x + y² + 4 – 4y = 13

x ² + y² -4x – 4y +8 -13 =0

x ² + y² -4x – 4y -5 =0

Q15.Does the point(-2.5,3.5) lie inside,outside or on the circle x² + y² = 25 ?

Ans. The given equation of the circle is x² + y² = 25

Comparing the given equation with the standard equation of the circle

(x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We get h =0,k =0 and r = 5

## Extract of the exercise 11.1 of the chapter 11-Conic Section

The centre of the circle is (0,0) and radius 5 indicates that the point (-2.5,3.5) lies inside the circle since the distance of the point from the centre(0,0) is less than 5 , the radius of the circle.

The circle is the most common geometrical figure in a range of all conic sections. The circle can be driven by slicing a plane parallel to its base and perpendicular to the axis of the cone.

### The standard equation of the circle

The standard equation of the circle  is

(x – h)² + (y – k)² = r²

Where (h,k) are the coordinates of the circle and r is the radius of the circle

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