NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 – Conic Sections

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 focuses entirely on circles, which is the first conic section introduced in the chapter. This exercise covers finding the equation of a circle when the centre and radius are given, converting the general form of a circle's equation into standard form by completing the square, and determining the centre and radius from a given equation. Students also learn to find the equation of a circle passing through given points with conditions on its centre.

Exercise 10.1 is an essential part of the CBSE Class 11 Maths syllabus 2025–26 and forms the foundation for understanding all conic sections. The questions in this exercise are straightforward yet conceptually important, and mastering them helps students tackle more complex problems in Exercise 10.2, 10.3, and 10.4. A solid grasp of circles is also beneficial for JEE Main and other competitive examinations.

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 – Conic Sections

Conic Sections

Medium

Q.

In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2
2. centre (–2,3) and radius 4
3. $\mathbf{centre}\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{)}\mathbf{and}\mathbf{ }\mathbf{radius}\mathbf{ }\frac{\mathbf{1}}{\mathbf{12}}$
4. $\mathbf{centre}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{and}\mathbf{ }\mathbf{radius}\mathbf{ }\sqrt{\mathbf{2}}$
5. $\mathbf{centre}\mathbf{(}\mathbf{-}\mathbf{a}\mathbf{,}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{radius}\mathbf{ }\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}$

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Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{major}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{minor}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{)}$

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Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:
$\mathbf{Vertices}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{7}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{e}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}$

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Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $\mathbf{Vertices}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{foci}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}$ 

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Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $$\mathbf{Vertices}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{foci}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{8}\mathbf{)}$$ 

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Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $$\mathbf{Vertices}\mathbf{\quad }\mathbf{(}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{foci}\mathbf{\quad }\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}$$ 

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Conic Sections

Medium

Q.

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

1.  $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{=}\mathbf{1}$

2.  $\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{27}}\mathbf{=}\mathbf{1}$
3. 9y² – 4x² = 36
4. 16 x² – 9y² = 576
5. 5y² – 9x² = 36
6.  49y² – 16x² = 784

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Conic Sections

Medium

Q.

Find the equation for the ellipse that satisfies the given condition:
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

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Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Foci}\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{a}\mathbf{=}\mathbf{4}$

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Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:A
$\mathbf{Length}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{minor}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{16}\mathbf{,}\mathbf{ }\mathbf{foci}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{6}\mathbf{)}$

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Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{major}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\sqrt{\mathbf{5}}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{minor}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$

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Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Vertices}\mathbf{(}\mathbf{\pm }\mathbf{6}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{foci}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$

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Conic Sections

Medium

Q.

In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)² + (y – 3)² = 36
7. x² + y² – 4x – 8y – 45 = 0
8. x² + y² – 8x + 10y – 12 = 0
9. 2x² + 2y² – x = 0

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Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Vertices}\mathbf{(}\mathbf{\pm }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{foci}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$

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Conic Sections

Medium

Q.

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

1. $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{36}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{=}\mathbf{1}$

2. $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{25}}\mathbf{=}\mathbf{1}$

3. $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{=}\mathbf{1}$

4. $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{25}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{100}}\mathbf{=}\mathbf{1}$

5. $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{49}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{36}}\mathbf{=}\mathbf{1}$

6. $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{100}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{400}}\mathbf{=}\mathbf{1}$
7. 36x² + 4y² = 144
8. 16x² + y² = 16
9. 4x² + 9y² = 36

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Conic Sections

Medium

Q.

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
7. Focus (6,0); directrix x = – 6
8. Focus (0,–3); directrix y = 3
9. Vertex (0,0); focus (3,0)
10. Vertex (0,0); focus (–2,0)
11. Vertex (0,0) passing through (2,3) and axis is along x-axis.
12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

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Conic Sections

Medium

Q.

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

1. y² = 12x
2. x² = 6y
3. y² = – 8x
4. x² = – 16y
5. y² = 10x
6. x² = – 9y

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Conic Sections

Easy

Q.

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

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Conic Sections

Easy

Q.

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

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Conic Sections

Difficult

Q.

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

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Conic Sections

Difficult

Q.

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

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Conic Sections

Difficult

Q.

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.

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Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:
$\mathbf{Foci}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\sqrt{\mathbf{10}}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{passing}\mathbf{ }\mathbf{through}\mathbf{ }\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{.}$

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Question 1. Find the equation of the circle with centre (0, 2) and radius 2.

The equation of a circle with centre (h, k) and radius r is: (x − h)² + (y − k)² = r²

Here, centre (h, k) = (0, 2) and radius r = 2.

(x − 0)² + (y − 2)² = 2² x² + y² + 4 − 4y = 4

Therefore, the equation of the circle is x² + y² − 4y = 0.

Question 2. Find the equation of the circle with centre (−2, 3) and radius 4.

The equation of a circle with centre (h, k) and radius r is: (x − h)² + (y − k)² = r²

Here, centre (h, k) = (−2, 3) and radius r = 4.

(x + 2)² + (y − 3)² = 4² x² + 4x + 4 + y² − 6y + 9 = 16

Therefore, the equation of the circle is x² + y² + 4x − 6y − 3 = 0.

Question 3. Find the equation of the circle with centre (1/2, 1/4) and radius 1/12.

The equation of a circle with centre (h, k) and radius r is: (x − h)² + (y − k)² = r²

Here, centre (h, k) = (1/2, 1/4) and radius r = 1/12.

(x − 1/2)² + (y − 1/4)² = (1/12)²

x² − x + 1/4 + y² − y/2 + 1/16 = 1/144

Multiplying throughout by 144 (LCM): 144x² − 144x + 36 + 144y² − 72y + 9 − 1 = 0 144x² + 144y² − 144x − 72y + 44 = 0

Dividing by 4:

Therefore, the equation of the circle is 36x² + 36y² − 36x − 18y + 11 = 0.

Question 4. Find the equation of the circle with centre (1, 1) and radius √2.

The equation of a circle with centre (h, k) and radius r is: (x − h)² + (y − k)² = r²

Here, centre (h, k) = (1, 1) and radius r = √2.

(x − 1)² + (y − 1)² = (√2)² x² − 2x + 1 + y² − 2y + 1 = 2

Therefore, the equation of the circle is x² + y² − 2x − 2y = 0.

Question 5. Find the equation of the circle with centre (−a, −b) and radius √(a² − b²).

The equation of a circle with centre (h, k) and radius r is: (x − h)² + (y − k)² = r²

Here, centre (h, k) = (−a, −b) and radius r = √(a² − b²).

(x + a)² + (y + b)² = (√(a² − b²))² x² + 2ax + a² + y² + 2by + b² = a² − b² x² + y² + 2ax + 2by + b² + b² = a²

Therefore, the equation of the circle is x² + y² + 2ax + 2by + 2b² = 0.

Question 6. Find the centre and radius of the circle (x + 5)² + (y − 3)² = 36.

The equation is (x + 5)² + (y − 3)² = 36.

⟹ {x − (−5)}² + (y − 3)² = 6²

This is of the form (x − h)² + (y − k)² = r², where h = −5, k = 3, r = 6.

Therefore, the centre is (−5, 3) and the radius is 6.

Question 7. Find the centre and radius of the circle x² + y² − 4x − 8y − 45 = 0.

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

This is of the form (x − h)² + (y − k)² = r², where h = 2, k = 4, r = √65.

Therefore, the centre is (2, 4) and the radius is √65.

Question 8. Find the centre and radius of the circle x² + y² − 8x + 10y − 12 = 0.

x² + y² − 8x + 10y − 12 = 0 ⟹ (x² − 8x) + (y² + 10y) = 12 ⟹ {x² − 2(x)(4) + 4²} + {y² + 2(y)(5) + 5²} − 16 − 25 = 12 ⟹ (x − 4)² + (y + 5)² = 53 ⟹ (x − 4)² + {y − (−5)}² = (√53)²

This is of the form (x − h)² + (y − k)² = r², where h = 4, k = −5, r = √53.

Therefore, the centre is (4, −5) and the radius is √53.

Question 9. Find the centre and radius of the circle 2x² + 2y² − x = 0.

2x² + 2y² − x = 0 ⟹ (2x² − x) + 2y² = 0 ⟹ 2[(x² − x/2) + y²] = 0 ⟹ {x² − 2x(1/4) + (1/4)²} + y² − (1/4)² = 0 ⟹ (x − 1/4)² + (y − 0)² = (1/4)²

This is of the form (x − h)² + (y − k)² = r², where h = 1/4, k = 0, r = 1/4.

Therefore, the centre is (1/4, 0) and the radius is 1/4.

Question 10. Find the equation of the circle passing through (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

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) lies on 4x + y = 16: 4h + k = 16 …(iii)

From (i) and (ii): 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)

Solving (iii) and (iv): h = 3, k = 4

Substituting in (i): (4 − 3)² + (1 − 4)² = r² 1 + 9 = r² ⟹ r² = 10

(x − 3)² + (y − 4)² = 10 x² − 6x + 9 + y² − 8y + 16 = 10

Therefore, the equation of the circle is x² + y² − 6x − 8y + 15 = 0.

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

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

Since the circle passes through (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) lies on x − 3y − 11 = 0: h − 3k = 11 …(iii)

From (i) and (ii): 4 − 4h + h² + 9 − 6k + k² = 1 + 2h + h² + 1 − 2k + k² 4 − 4h + 9 − 6k = 2 + 2h − 2k 6h + 4k = 11 …(iv)

Solving (iii) and (iv): h = 7/2, k = −5/2

Substituting in (i): (2 − 7/2)² + (3 + 5/2)² = r² (−3/2)² + (11/2)² = r² 9/4 + 121/4 = r² r² = 130/4

(x − 7/2)² + (y + 5/2)² = 130/4

Expanding and simplifying:

Therefore, the equation of the circle is x² + y² − 7x + 5y − 14 = 0.

Question 12. Find the equation of the circle with radius 5 whose centre lies on the x-axis and passes through the point (2, 3).

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

Since the centre lies on the x-axis, k = 0, and r = 5.

Since the circle passes through (2, 3): (2 − h)² + 3² = 25 (2 − h)² = 16 (2 − h) = ±4

Case 1: 2 − h = 4 ⟹ h = −2 Case 2: 2 − h = −4 ⟹ h = 6

When h = −2: (x + 2)² + y² = 25 x² + y² + 4x − 21 = 0

When h = 6: (x − 6)² + y² = 25 x² + y² − 12x + 11 = 0

Therefore, the equation of the circle is x² + y² + 4x − 21 = 0 or x² + y² − 12x + 11 = 0.

Question 13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

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

Since the circle passes through (0, 0): h² + k² = r²

The circle makes intercept a on x-axis ⟹ passes through (a, 0): (a − h)² + k² = h² + k² a² − 2ah = 0 a(a − 2h) = 0 Since a ≠ 0, h = a/2

The circle makes intercept b on y-axis ⟹ passes through (0, b): h² + (b − k)² = h² + k² b² − 2bk = 0 b(b − 2k) = 0 Since b ≠ 0, k = b/2

Equation of the required circle: (x − a/2)² + (y − b/2)² = (a/2)² + (b/2)² x² − ax + a²/4 + y² − by + b²/4 = a²/4 + b²/4

Therefore, the equation of the circle is x² + y² − ax − by = 0.

Question 14. Find the equation of a circle with centre (2, 2) and passing through the point (4, 5).

Centre (h, k) = (2, 2). Since the circle passes through (4, 5), the radius is: r = √((2 − 4)² + (2 − 5)²) = √(4 + 9) = √13

Equation of the circle: (x − 2)² + (y − 2)² = (√13)² x² − 4x + 4 + y² − 4y + 4 = 13

Therefore, the equation of the circle is x² + y² − 4x − 4y − 5 = 0.

Question 15. Does the point (−2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?

The given circle: x² + y² = 25 ⟹ (x − 0)² + (y − 0)² = 5²

Centre = (0, 0) and radius = 5.

Distance between point (−2.5, 3.5) and centre (0, 0): d = √((−2.5)² + (3.5)²) = √(6.25 + 12.25) = √18.25 ≈ 4.3

Since d ≈ 4.3 < 5 (radius), the point lies inside the circle.

Therefore, the point (−2.5, 3.5) lies inside the circle.

Frequently Asked Questions (FAQs)

Q1. What is the standard equation of a circle? The standard equation of a circle with centre (h, k) and radius r is (x − h)² + (y − k)² = r². When the centre is at the origin, it simplifies to x² + y² = r².

Q2. How do you find the centre and radius from the general equation of a circle? The general form is x² + y² + 2gx + 2fy + c = 0. Complete the square for both x and y terms to convert it to standard form. The centre is (−g, −f) and the radius is √(g² + f² − c).

Q3. How do you determine if a point lies inside, outside, or on a circle? Calculate the distance d from the point to the centre. If d < r, the point is inside; if d = r, it is on the circle; if d > r, it is outside the circle.

Q4. How do you find the equation of a circle passing through three given points? Substitute all three points into the general equation x² + y² + Dx + Ey + F = 0 to get three equations. Solve these simultaneously to find D, E, and F.

Q5. How many questions are in NCERT Class 11 Maths Chapter 10 Exercise 10.1? Exercise 10.1 contains 15 questions, all based on circles — including finding the equation of a circle from given centre and radius, finding centre and radius from a given equation, and finding equations of circles satisfying special conditions.