## Maths NCERT Solutions Class 11 | CBSE Class 11th Mathematics

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__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 circleand the distance from the centre to any point on the circle is called the** radius** of the circle.

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}= r^{2}**

- The general equation of the circle is given by
**x**, where^{2}+ y^{2}+ 2gx + 2fy + c = 0*g*,*f*and*c*

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

(b) The radius of the circle is

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**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} = 3^{2}

or**(x-1) ^{2} + (y-2)^{2}=9**

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**Question: **Find the centre and radius of the circle *x*^{2} + *y*^{2} -2*x*+ 4*y *– 8 = 0.

**Solution: **First of all,we write the given equation in the form (*x*^{2} – 2*x*) + ( *y*^{2} + 4*y*) = 8

Now, completing the squares, we get

(*x*^{2} – 2*x *+ 1) + ( *y*^{2} + 4*y *+ 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 x^{2}+ y^{2} + 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

**x**^{2}**+ y ^{2}-6x-8y+15 = 0**

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