CBSE Class 11 maths NCERT solutions Circle

CBSE Class 11 maths NCERT solutions Circle

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 circle and the distance from the centre to any point on the circle is called the radius of the circle.

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

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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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