Videos Related with Circle

Note on Circle

  • Note
  • Things to remember
  • Videos
  • Exercise
  • Quiz

Circle

A circle is a closed curve such that every point on the curve is at a constant distance from a fixed point. Circle may also be defined as a locus of a point which moves so that its distance from a fixed point is constant. The fixed point is called the centre and the constant distance is called the radius of the circle.

Equation of a circle

1. Centre at the origin (Standard form)
Let O (0, 0) be the centre and r be the radius of the circle.
Let P (x, y) be any point on the circle.
Then, OP = r
Squaring both sides we have,
OP2= r2
or, (x - 0)2 + (y - 0)2 = r2
or, x2 + y2 = r2
This relation is true for any point P (x , y) on the circle. So, it is the equation of the circle.

Centre at the origin (Standard form)
Centre at the origin (Standard form)

2. Centre at any point (Central form)
Let Q (h, k) be the centre and r be the radius of the circle.
Let P (x, y) be any point on the circle.
Then, QP = r
Squaring both sides we have,
QP2 = r2
or, (x - h)2 + (y k)2 = r2
This is the equation of the circle.

Centre at any point (Central form
Centre at any point (Central form

3. Circle with a given diameter (Diameter form)
Let A (x1, y1) and B(x2, y2) be the ends of a diameter of a circle.
Let P (x, y) be any point on the circle. Join AP, BP and AB.
Since AB is a diameter of the circle, then \(\angle\)APB is a right angle.
Now,
Slope of AP = \(\frac {y - y_1}{x - x_1}\)
Slope of BP = \(\frac {y - y_2}{x - x_2}\)
Since,
AP is perpendicular to BP, the product of their slopes must be -1.
Hence,
\(\frac {y - y_1}{x - x_1}\) . \(\frac {y - y_2}{x - x_2}\) = -1
or, (y - y1) . (y - y2) = - (x - x1) . (x - x2)
or, (x - x1) . (x - x2) +(y - y1) . (y - y2) = 0
This relation is satisfied by any point on the circle. So, it is the equation of the circle.

Circle with a given diameter (Diameter form)
Circle with a given diameter (Diameter form)

4. General equation of the circle
Let Q (h, k) be the centre and r be the radius of the circle.
Let P (x, y) be any point on the circle.
Then,
QP = r
Squaring on both sides,
QP2 = r2
or, (x - h)2 + (y - k)2 = r2
or, x2 - 2hx + h2 + y2 - 2ky + k2 = r2
or, x2 + y2 - 2hx - 2ky + h2 + k2 - r2 = 0
This is the equation of a circle having centre at the point (h, k) and radius r.
Putting -2h = 2g, -2k = 2f and h2 + k2 - r2 = c we have,
x2 + y2 + 2gx + 2fy + c = 0
This is general equation of the circle.

General equation of the circle
General equation of the circle

Note: The general equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0.
Comparing this equation with the general equation of second degree
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 we have,
- coefficient of x2 and y2 are equal
- coefficient of xy is zero
Hence,
The general equation of second degreeax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a circle if
- coefficients of x2 and y2 are equal i.e. a = b
- coefficient of xy is zero i.e. h = 0

Centre and radius of a circle

The general equation of a circle is:

x2 + y2 + 2gx + 2fy + c = 0

or, x2 + 2gx + y2+ 2fy = -c

or, x2 + 2gx + g2 + y2 + 2fy + f2 = g2 + f2 - c

or, (x + g)2 + (y + f)2 = (\(\sqrt {g^2 + f^2 - c}\))2

Comparing this equation with (x - h)2 + (y - k)2 = r2, we have

h = -g, k = -f and r = \(\sqrt {g^2 + f^2 - c}\)

Hence,

Centre of the circle (h, k) = (-g, -f)

Radius of the circle (r) = \(\sqrt {g^2 + f^2 - c}\)

Equation of a circle passing through three points

Equation of a circle passing through three point
Equation of a circle passing through three point

Let, A (x1, y1), B (x2, y2) and C (x3, y3) be three points of a circle.

Let, P (h, k) be the centre of the circle.

Then,

AP = BP = CP

or, AP2 = BP2 = CP2

By using distance formula,

We have,

AP2= (x1 - h)2 + (y1 - k)2

BP2= (x2 - h)2 + (y2 - k)2

CP2= (x3 - h)2 + (y3 - k)2

Taking AP2 = BP2 we have,

(x1 - h)2 + (y1- k)2 = (x2 - h)2 + (y2 - k)2 .....................................................(i)

Taking BP2 = CP2 we have,

(x2 - h)2 + (y2- k)2 = (x3 - h)2 + (y3 - k)2 .....................................................(ii)

By solving (i) and (ii) we will get the values of h and k. Hence, we get centre P (h, k) of the circle. Then length of AP or BP or CP gives the radius r of the circle.

Now, Putting the values of h, k and r in (x - h)2 + (y - k)2 = r2 we get the required equation of the circle.

Equation of a circle in particular cases

1. When a circle touches the X - axis

Let the centre of a circle be C(h, k) and radius r. If this circle touches X- axis and the circle is in the first or second quadrant, then r = k.
If the circle is in the third or fourth quadrant, then r = -k.
Now,
The equation of a circle touching the X- axis is
(x - h)2 + (y - k)2 = r2
or, (x - h)2 + (y - k)2= k2

When a circle touches the X - axis
When a circle touches the X - axis

2. When a circle touches the Y- axis

Let C (h, k) and r be the centre and radius of a circle.
If this circle touches the Y- axis and it is in the first or fourth quadrant, then r = h.
If the circle is in the second or third quadrant, then r = -h.
Now,
The equation of a circle touching the Y- axis is
(x - h)2 + (y - k)2 = r2
or,(x - h)2 + (y - k)2 = h2

When a circle touches the Y- axis
When a circle touches the Y- axis

3. When a circle touches both the positive axes

Let C(h, k) and r be the centre and radius of a circle, if this circle touches both the positive axes, then h = k = r.
Now,
The equation of the circle is:
(x - h)2 + (y - k)2 = r2
or,(x - h)2 + (y - k)2 = h2
or,(x - k)2 + (y - h)2 = k2
or,(x - r)2 + (y - k)2 = k2

When a circle touches both the positive axes
When a circle touches both the positive axes



Equation of a circle

  • Centre at the origin (Standard form)
  • Centre at any point (Central form)
  • Circle with a given diameter (Diameter form)
  • General equation of the circle

Equation of a circle in particular cases

  • When a circle touches the X - axis
  • When a circle touches the Y- axis
  • When a circle touches both the positive axes
.

Very Short Questions

Here,

2x2 + 2y2 - 5x - 7y - 23 = 0

or, \(\frac {2x^2}2\) + \(\frac {2y^2}2\) - \(\frac 52\)x - \(\frac 72\)y - \(\frac {23}2\) = 0

or, x2 - \(\frac 52\)x + y2 - \(\frac 72\)y - \(\frac {23}2\) = 0

or, x2 - 2.x.\(\frac 54\) + (\(\frac 54)\)2 + y2 - 2.y.\(\frac 74\) + (\(\frac 74\))2 + (\(\frac 74\))2- (\(\frac 74\))2 - \(\frac {23}2\) = 0

or, (x - \(\frac 54\))2 + (y - \(\frac 74\))2 = \(\frac {25}{16}\) + \(\frac {49}{16}\) + \(\frac {23}2\) = \(\frac {25 + 49 + 184}{16}\)

or, (x - \(\frac 54\))2 + (y - \(\frac 74\))2 = \(\frac {258}{16}\)............................(1)

Eqn of circle is: (x - h)2+ (y - x)2 = r2......................(2)

Comparing (1) and (2)

Centre of circle (h, k) = (\(\frac 54\), \(\frac 74\))

and radius of circle (r) = \(\frac {\sqrt {258}}4\) units Ans

Here,

x2 + y2 - 20y + 75 = 0

or, x2 + y2 - 2.y.10 + (10)2 - (10)2 + 75 = 0

or, (x - 0)2 + (y - 10)2- 100 + 75 = 0

or, (x - 0)2 + (y - 10)2- 25 = 0

or, (x - 0)2 + (y - 10)2= 25

or, (x - 0)2 + (y - 10)2=(5)2............................(1)

The eqn of circle is: (x - h)2 + (y - k)2 = r2.....................(2)

Comparing (1) and (2)

Centre of circle (h, k) = (0. 10) Ans

Here,

x2 + y2 - 4x + 10y - 7 = 0

or, x2- 4x + y2 + 10y - 7 = 0

or, x2 - 2.x.2 + 22 - 22 + y2 + 2.y.5 + 52 - 52 - 7 = 0

or, (x - 2)2 + (y + 5)2 - 4 - 25 - 7 = 0

or, (x - 2)2 + (y + 5)2 - 36 = 0

or, (x - 2)2 + (y + 5)2= 36

or, (x - 2)2 + (y + 5)2= 62...........................(1)

The eqn of the circle is: (x - h)2+ (y - k)2 = r2.........................(2)

Comparing (1) and (2)

The length of radius of a circle is: 6 units Ans

Here,

x2 + y2 + 4x - 6y + 4 = 0

or, x2 + 4x+ 4 + y2 - 6y = 0

or, x2 + 2.x.2 + 22+ y2 - 2.y.3 + 32 - 32= 0

or, (x + 2)2 + (y - 3)2 = 32.....................................(1)

The eqn of circle is: (x - h)2 + (y - k)2 = r2...........................(2)

Comparing (1) and (2)

The length of radius of the circle (r) = 3 unitsAns

Here,

Centre of circle (h, k) = (3, 0)

Radius of circle (r) = 5 units

The equation of circle is: (x - h)2 + (y - k)2 = r2

(x - 3)2 + (y - 0)2 = 52

or, x2 - 6x + 9 + y2 = 25

or, x2 + y2 - 6x + 9 - 25 = 0

∴x2 + y2 - 6x -16 = 0Ans

Here,

x2 + y2 + 4x - 4y -1 = 0

or, x2 + 4x + y2 - 4y - 1 = 0

or, x2 + 2.x.2 + 22 - 22 + y2 - 2.y.2 + 22 - 22 - 1 = 0

or, (x + 2)2 + (y - 2)2 - 9 = 0

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

or, (x + 2)2 + (y - 2)2=32................................(1)

Equation of circle, (x - h)2 + (y - k)2 = r2...................................(2)

Comparing (1) and (2)

h = -2

k = 2

r = 3

∴ Centre = (-2, 2) and radius = 3 unitsAns

Here,

x2+ y2 - 2y = 24

or, x2 + y2 - 2.y.1 + 12- 12 = 24

or, x2 + (y - 1)2 - 1 = 24

or, x2 + (y - 1)2 = 25

or, x2 + (y - 1)2 = 52.............................(1)

The equation of circle is: (x - h)2 + (y - k)2 = a2........................(2)

Comparing (1) and (2)

(h, k) = (0, 1)

a = 5

Hence, centre = (0, 1) and radius = 5 unitsAns

Given eqn is:

(x + 5)2 + y2 = 121

or, (x + 5)2 + (y - 0)2 = (11)2..........................(1)

The eqn of the circle is:

(x - h)2 + (y - k)2 = r2..............................(2)

Comparing (1) and (2)

h = -5

k = 0

r = 11

∴ The centre of circle (h, k) = (-5, 0) and radius (r) = 11

∴ Diameter = 2r = 2× 11 = 22 unitsAns

Here,

x2 + y2 - 4x - 6y - 12 = 0

or, x2 - 4x + y2 - 6y - 12 = 0

or, x2 - 2.x.2 + 22 - 22 + y2- 2.y.3 + 32- 32- 12 = 0

or, (x - 2)2 + (y - 3)2 - 12 - 4 - 9= 0

or, (x - 2)2 + (y - 3)2 - 25 = 0

or, (x - 2)2 + (y - 3)2 = 25

or, (x - 2)2 + (y - 3)2 = 52............................(1)

The equation of circle is: (x - h)2 + (y - k)2 = r2.............................(2)

Comparing (1) and (2)

(h, k) = (3, 2) and r = 5

∴ The centre of circle = (3, 2) and radius of circle (r) = 5 unitsAns

Here,

(x1 - y1) = (2, 4)

(x2 - y2) = (3, -6)

The equation of a circle in diameter form:

(x - x1) (x - x2) + (y - y1) (y - y2) = 0

or, (x - 2) (x - 3) + (y - 4) (y + 6) = 0

or, x2 - 3x - 2x + 6 + y2 + 6y - 4y - 24 = 0

or, x2 - 5x + 6 + y2 + 2y - 24 = 0

∴ x2 + y2 - 5x + 2y - 18 = 0Ans

z

Let: A(4, 2) and B(3, 5) are the two ends of the diameter of a circle.

The equation of a circle in diameter form:

(x - x1) (x - x2) + (y - y1) (y - y2) = 0

or, (x, 4) (x, 3) + (y - 2) (y - 5) = 0

or, x2 - 3x - 4x + 12 + y2 - 5y - 2y + 10 = 0

∴ x2 + y2 - 7x - 7y + 22 = 0Ans

Here,

Radius of circle (r) = 5 units

Eqn of two diameters are:

x = 3y.............................(1)

y = 2...............................(2)

Putting the value of y in eqn (1)

x = 3× 2 = 6

Intersection point of the two diameters is centre of circle so:

Centre of circle = (h, k) = (6, 2)

Eqn of circle is:

(x - h)2 + (y - k)2 = r2

or, (x - 6)2 + (y - 2)2 = 52

or, x2- 12x + 36 + y2 - 4y + 4 - 25 = 0

∴ x2 + y2 - 12x - 4y + 15 = 0Ans

Here,

Centre of circle (h, k) = (-2, 3)

Radius (r) = k = 3

The eqn of circle is:

(x - h)2 + (y - k)2 = r2

or, (x + 2)2 + (y - 3)2 = 32

or, x2 + 4x + 4 + y2 - 6y + 9 = 9

or, x2 + y2+ 4x - 6y + 13 - 9 = 0

∴ x2 + y2+ 4x - 6y + 4 = 0Ans

Here,

Centre of circle (h, k) = (4, -3)

Radius (r) = h = 4

The equation of circle is:

(x - h)2 + (y - k)2 = r2

or, (x - 4)2 + (y + 3)2 = 42

or, x2 - 8x + 16 + y2 + 6y + 9 = 16

or, x2 + y2 - 8x + 6y + 25 = 16

or, x2 + y2 - 8x + 6y + 25 -16 = 0

∴ x2 + y2 - 8x + 6y + 9 = 0Ans

Here,

Radius of circle (r) = 5 units

h = k = r = 5 [\(\because\) touches on both axis]

The eqn of circle is:

(x - h)2 + (y - k)2 = r2

or, (x - 5)2 + (y - 5)2 = 52

or, x2 - 10x + 25 + y2 - 10y + 25 = 25

or, x2 + y2 - 10x - 10y + 50 - 25 = 0

∴ x2 + y2 - 10x - 10y + 25 = 0Ans

Here,

Centre of circle = (h, k) = (6, 8)

The eqn of circle is:

(x - h)2 + (y - k)2 = r2

or, (x - 6)2 + (y - 8)2 = r2..............................(1)

The eqn (1) passes through the point (0, 0)

(0 - 6)2+ (0 - 8)2 = r2

or, 36 + 64 = r2

or, 100 = r2

or, (r)2 = (10)2

∴ r = 10 units

Putting the value of r in eqn (1)

(x - 6)2 + (y - 8)2 = (10)2

or, x2- 12x + 36 + y2 - 16y + 64 = 100

or, x2 + y2 - 12x - 16y + 100 = 100

∴x2 + y2 - 12x - 16y= 0Ans

0%
  • x+ y- 20y + 75 = 0

    (10 , 20)


    (10 , 20)


    (0 , 10)


    (0 , 0)


  • x2 + y2 + 4x - 5 = 0 

    (2 , 0)


    (0 , 2)


    (0)


    (-2 , 0)


  • x2 + y2 - 6x + 2y + 1  = 0

    (-1 , -2)


    (-1 , -1)


    (2 , -1)


    (3 , -1)


  • Find the length and radius of the circle . 


    x2 + y2 + 4x - 6y + 4 = 0

    16 units


    5 units


    3 units


    24 units


  • Find the length of the circumference of a circle having the equation x2 + y2- 2y-48 = 0

    96 units


    98 units


    24 units


    44 units


  • Find the equation of the circle having centren (0 , 0) and radius 3 units . 

    x2 - y2 = 10


    x2 - y2 = 9


    x2 + 9 = y2


    x2 + y2 = 9


  • Find the equation of the circle with centre (2 , -1) and radius 3 units . 

    x2 + y2 - 4x + 2y = 4


    x2 + y2 + 2x + 4y = 4


    x2 - y2 + 4x - 2y = -4


    x2 + y2 + 4x + 2y = -4


  • Find the equation of a circle having centre (1 , -2) and the radius 2 (sqrt{5}) units . 

    x2 - y2 - 4x + 2y = 0


    x2 + y2 - 2x + 4y = 0


    x2 - y+ 2x - 4y = 0


    x2 - y2 + 2x + 4y = 0


  • What is the length of the radius of the circle having centre at P (3 , 4) and that touches the X-axis at (3 , 0) ?

    4 units


    3 units


    16 units


    2 units


  • Find the equation of the circle having centre (3 , 6) and touching the x-axis. 

    x2 - y2 + 6x - 12 + 9y = 0


    x + y - 6x - 12y + 9 = 0


    x2 + y2 -6x - 12y + 9 = 0


    x2 - y2 -6x + 11y + 9 = 0


  • Given points are  the end of the circle . Find the equation of the circle.

    A(5 , 6) and B(3 , 4)

    x2 - y2 + 8x + 10y - 39 = 0


    x2 + y2 - 8x - 10y + 39 = 0


    x2 + y2 + 8x + 10y - 39 = 0


    x2 + y2 + 8x - 10y + 0 = 39


  • Given points are  the end of the circle . Find the equation of the circle.

    (1 , 2) and (3 , 6)

    x2 + y2 - 4x - 8y + 15  = 0


    x2 - y2 + 4x + 8y + 0  = 15


    x2 - y2 + 4x + 8y - 15  = 0


    x2 - y2 - 4x - 8y - 15  = 0


  • Given points are  the end of the circle . Find the equation of the circle.

    ( -1 , 0) and (7 , 4). 

    x2 - y2 - 6x -4y - 7 = 0


    x2 + y2 -6x -4y - 7 = 0


    x2 - y2 -6x - 6y - 7 = 0


    x2 + y2 -6x + 4y + 0 = 7


  • Find the coordinates of centre and radius of the circle. 

    x2 + y2 - 2x - 6y +  1 = 0

    (1 , 3)  , 3 units


    4 , (1 , 3) units


    (3 , 1) , 4 units


    1  ,(4 , 3) units


  • Find the coordinates of centre and radius of the circle. 
    2x - 6y - x2 - y= 1

    (1 , -3) , 3 units


    (-1 , -3) , 1 units


    (3 , 1)  units


    (1 , 3) , 3 units


  • You scored /15


    Take test again

DISCUSSIONS ABOUT THIS NOTE

You must login to reply

Forum Time Replies Report


You must login to reply

Sagunsharma

Find the equation of circle passing through the point (6,-1) and touch the y axis at (0,5)


You must login to reply

Rachana

If the centre of circle x^2 y^2 - ax - by - 12 = 0 is (2,3),find the values of a and b.


You must login to reply

Sudip kunwar

If a line x y=1 cuts a circle x^2 y^2=1 at two points ,find the distance between two points.


You must login to reply