Notes on Circle | Grade 10 > Optional Mathematics > Co-ordinate Geometry | KULLABS.COM

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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.

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.

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.

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.

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

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

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

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

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
.

### Questions and Answers

#### Click on the questions below to reveal the answers

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

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%

(0 , 0)

(10 , 20)

(0 , 10)

(10 , 20)

(0)

(2 , 0)

(-2 , 0)

(0 , 2)

(-1 , -2)

(-1 , -1)

(3 , -1)

(2 , -1)

3 units

16 units

5 units

24 units

96 units

44 units

98 units

24 units

x2 - y2 = 9

x2 + y2 = 9

x2 + 9 = y2

x2 - y2 = 10

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

x2 + y2 - 4x + 2y = 4

x2 + y2 + 4x + 2y = -4

x2 - y2 + 4x - 2y = -4

x2 + y2 + 2x + 4y = 4

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

x2 - y2 + 2x + 4y = 0

x2 - y2 - 4x + 2y = 0

x2 + y2 - 2x + 4y = 0

x2 - y+ 2x - 4y = 0

4 units

16 units

3 units

2 units

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

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

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

x + y - 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 + 0 = 39

x2 - y2 + 8x + 10y - 39 = 0

• ### 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 + 0 = 7

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

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

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

• ### Find the coordinates of centre and radius of the circle. x2 + y2 - 2x - 6y +  1 = 0

(1 , 3)  , 3 units

1  ,(4 , 3) units

4 , (1 , 3) units

(3 , 1) , 4 units

• ### Find the coordinates of centre and radius of the circle. 2x - 6y - x2 - y2 = 1

(1 , -3) , 3 units

(-1 , -3) , 1 units

(3 , 1)  units

(1 , 3) , 3 units

## ASK ANY QUESTION ON Circle

Forum Time Replies Report

##### Sagunsharma

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

##### Rachana

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