Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections (Exercise 10.1) with step-by-step explanations as per the latest CBSE and NCERT syllabus (2025–2026). This exercise introduces students to the fundamental concepts of conic sections, including circle, ellipse, parabola, and hyperbola, along with their standard equations and geometric properties. Each question is solved clearly with formulas, derivations, and diagrams to help students strengthen their understanding of coordinate geometry. These detailed NCERT solutions are ideal for CBSE Class 11 exams and for competitive exams like JEE Main, JEE Advanced, NDA, and CUET. Click the print button to download study material and notes in PDF format.
NCERT Question.1 : Find the equation of the circle with centre (0, 2) and radius 2.
Solution:
The equation of a circle with centre $(h, k)$ and radius $r$ is
$$(x – h)^2 + (y – k)^2 = r^2$$
Here, centre $(h, k) = (0, 2)$ and radius $r = 2$.
So, the equation of the circle is $(x – 0)^2 + (y – 2)^2 = 2^2$.
Expanding using $(a – b)^2 = a^2 – 2ab + b^2$:
$$x^2 + y^2 – 4y + 4 = 4$$
$$x^2 + y^2 – 4y = 0$$
Equation of the circle: $x^2 + y^2 – 4y = 0$
NCERT Question.2 : Find the equation of the circle with centre (-2, 3) and radius 4.
Solution:
Centre $(h, k) = (-2, 3)$, radius $r = 4$.
Equation: $(x + 2)^2 + (y – 3)^2 = 4^2$
Expanding: $x^2 + 4x + 4 + y^2 – 6y + 9 = 16$
Simplifying: $x^2 + y^2 + 4x – 6y – 3 = 0$
Equation of the circle: $x^2 + y^2 + 4x – 6y – 3 = 0$
NCERT Question.3 : Find the equation of the circle with centre (1/2, 1/4) and radius 1/12.
Solution:
Centre $(h, k) = (1/2, 1/4)$, radius $r = 1/12$.
Equation: $(x – 1/2)^2 + (y – 1/4)^2 = (1/12)^2$
Expanding: $x^2 – x + 1/4 + y^2 – y/2 + 1/16 = 1/144$
Multiply through by 144 to remove denominators:
$144x^2 – 144x + 36 + 144y^2 – 72y + 9 – 1 = 0$
$144x^2 – 144x + 144y^2 – 72y + 44 = 0$
Divide by 4: $36x^2 – 36x + 36y^2 – 18y + 11 = 0$
Equation of the circle: $36x^2 + 36y^2 – 36x – 18y + 11 = 0$
NCERT Question.4 : Find the equation of the circle with centre (1, 1) and radius √2.
Solution:
Centre $(h, k) = (1, 1)$, radius $r = \sqrt{2}$.
Equation: $(x – 1)^2 + (y – 1)^2 = (\sqrt{2})^2$
Expanding: $x^2 – 2x + 1 + y^2 – 2y + 1 = 2$
Simplifying: $x^2 + y^2 – 2x – 2y = 0$
Equation of the circle: $x^2 + y^2 – 2x – 2y = 0$
NCERT Question.5 : Find the equation of the circle with centre (-a, -b) and radius √(a2 – b2).
Solution:
Centre $(h, k) = (-a, -b)$, radius $r = \sqrt{a^2 – b^2}$.
Equation: $(x + a)^2 + (y + b)^2 = (\sqrt{a^2 – b^2})^2$
Expanding: $x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 – b^2$
Simplifying: $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$
Equation of the circle: $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$
Download detailed NCERT solutions by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.
NCERT Question.6: Find the Centre and Radius of the Circle $(x + 5)^2 + (y – 3)^2 = 36$
Solution:
Given equation: $(x + 5)^2 + (y – 3)^2 = 36$
This can be written as $(x – (-5))^2 + (y – 3)^2 = 6^2$.
The general equation of a circle is $(x – h)^2 + (y – k)^2 = r^2$, where $h$, $k$, and $r$ are the centre coordinates and radius respectively.
Here, $h = -5$, $k = 3$, and $r = 6$.
Therefore, the centre is $(-5, 3)$ and its radius is $6$.
NCERT Question.7: Find the Centre and Radius of the Circle $x^2 + y^2 – 4x – 8y – 45 = 0$
Solution:
Given equation: $x^2 + y^2 – 4x – 8y – 45 = 0$
Rearranging: $(x^2 – 4x) + (y^2 – 8y) = 45$
Completing the squares:
$(x^2 – 2(2)x + 2^2) + (y^2 – 2(4)y + 4^2) – 4 – 16 = 45$
$(x – 2)^2 + (y – 4)^2 = 65$
This can be written as $(x – 2)^2 + (y – 4)^2 = (\sqrt{65})^2$.
So, $h = 2$, $k = 4$, and $r = \sqrt{65}$.
Therefore, the centre is $(2, 4)$ and its radius is $\sqrt{65}$.
NCERT Question.8: Find the Centre and Radius of the Circle $x^2 + y^2 – 8x + 10y – 12 = 0$
Solution:
Given equation: $x^2 + y^2 – 8x + 10y – 12 = 0$
Rearranging: $(x^2 – 8x) + (y^2 + 10y) = 12$
Completing the squares:
$(x^2 – 2(4)x + 4^2) + (y^2 – 2(-5)y + (-5)^2) – 16 – 25 = 12$
$(x – 4)^2 + (y + 5)^2 = 53$
This can be written as $(x – 4)^2 + (y – (-5))^2 = (\sqrt{53})^2$.
So, $h = 4$, $k = -5$, and $r = \sqrt{53}$.
Therefore, the centre is $(4, -5)$ and its radius is $\sqrt{53}$.
NCERT Question.9: Find the Centre and Radius of the Circle $2x^2 + 2y^2 – x = 0$
Solution:
Given equation: $2x^2 + 2y^2 – x = 0$
Dividing throughout by $2$: $x^2 + y^2 – \dfrac{x}{2} = 0$
Rewriting: $(x^2 – \frac{x}{2}) + y^2 = 0$
Completing the square for $x$:
$(x^2 – 2(\frac{1}{4})x + (\frac{1}{4})^2) + y^2 – (\frac{1}{4})^2 = 0$
$(x – \frac{1}{4})^2 + y^2 = (\frac{1}{4})^2$
Hence, $(x – \frac{1}{4})^2 + (y – 0)^2 = (\frac{1}{4})^2$.
So, $h = \frac{1}{4}$, $k = 0$, and $r = \frac{1}{4}$.
Therefore, the centre is $(\frac{1}{4}, 0)$ and its radius is $\frac{1}{4}$.
Download detailed NCERT solutions by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.
