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Maths NCERT Solutions Class 11 | CBSE Class 11th Mathematics

Maths NCERT Solutions Class 11 CBSE Class 11th Mathematics,CBSE Class 11th Mathematics,class 11 maths,class 11 maths NCERT solutions,NCERT maths solutions class 11

Maths NCERT Solutions Class 11 | CBSE Class 11th Mathematics

Maths NCERT Solutions Class 11 : Our main objective is to offer quality education to our students who are pursuing higher studies. We have experienced faculty members who leave no stone unturned in helping the students to score the best. We have prepared Class 11th online classes which cater to all your needs to score good marks in CBSE exams. Our online classes for Class 11 is unique & latest content is added regularly keeping in mind the latest pattern of question papers in CBSE board exams. You can also take our CBSE NCERT solutions of Maths and Science with detailed solutions & analysis in order to keep track with the pace of other students.

Class 11 maths ncert solutions – Circle

Circle:  A circle is the set of all points in a plane which is at a fixed distance from a fixed point in the plane. The fixed point is called the centre of the circleand the distance from the centre to any point on the circle is called the radius of the circle.
Maths NCERT Solutions Class 11 CBSE Class 11th Mathematics

In this figure,C is the centre and CP is a radius.


  • The equation of a circle with radius r having centre (h, k) is given by

(x – h)2 + (y – k)2= r2


  • The general equation of the circle is given by x2+ y2 + 2gx + 2fy + c = 0, where g, f and c

(a) The centre of this circle is (– g, – f)


(b) The radius of the circle is



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


Solution: We have   h = 1 ,k = 2 and r = 3

So equation of the circle is (x-1)2 + (y-2)2 = 32

or(x-1)2 + (y-2)2=9


For NCERT maths solutions class 11 practice sets of Conic Sections & more, Click at Class 11th Maths online classes 


Question: Find the centre and radius of the circle x2 + y2 -2x+ 4y – 8 = 0.


Solution: First of all,we write the given equation in the form (x2 – 2x) + ( y2 + 4y) = 8

Now, completing the squares, we get

(x2 – 2x + 1) + ( y2 + 4y + 4) = 8 + 1 + 4

(x – 1)2+ (y + 2)2 = 13

Comparing it with the standard form of the equation of the circle, we see that the

centre of the circle is (1, –2) and radius is  .


Alternative Method:

Comparing it with the general equation of the circle , we have

2g = -2 or g = -1

2f = 4 or f = 2

c = -8

So centre of circle is (1,-2) and

Radius is   =   .



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


Solution: Let the equation of circle is x2+ y2 + 2gx + 2fy + c = 0

Since it passes through (4,1) so it satisfies the equation of circle

ie 16+1+8g+2f+c = 0

or 8g+2f+c = -17                     _ (1)


Also it passes through (6,5) so we get

12g+10f+c = -61                       _(2)


Now centre is on line 4x+ y = 16 and we know centre is (-g,-f)

So it must satisfies the equation of line

ie -4g-f = 16                                   _(3)

Solving the above three equations,we get g =-3, f = – 4 and c = 15

Hence the equation of circle is

x2+ y2-6x-8y+15 = 0


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