Circle

Circle

Circle
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


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

z

Here,

AB is the diameter whose co-ordinates areA(x1, y1) and B(x2, y2).

Let: P(x, y) be any point on the circle.

Slope of PA = \(\frac {y - y_1}{x - x_1}\)

Slope of PB = \(\frac {y - y_2}{x - x_2}\)

\(\angle\)APB = 90° [\(\because\) The angle made in a semi-circle is a right angle.]

Now,

slope of PA× slope of PB = -1

or,\(\frac {y - y_1}{x - x_1}\)× \(\frac {y - y_2}{x - x_2}\) = -1

or, (y - y1) (y - y2) = - (x - x1) (x - x2)

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

Here,

Given points of the circle are:P(2, -2), Q(6, 6) and R(5, 7)

Let, centre of circle is: (h, k) and radius of circle is: a.

Equation of circle is (x - h)2 + (y - k)2 = a2...........................(1)

Given points P,Q and R passes through the equation (1)

(2 - h)2 + (-2, - k)2= a2.........................(2)

(6 - h)2 + (6 - k)2 = a2...........................(3)

(5 - h)2 + (7 - k)2 = a2...........................(4)

From equation (2) and (3)

(2 - h)2 + (-2, - k)2=(6 - h)2 + (6 - k)2

or, 4 - 4h + h2 + 4 + 4k + k2 = 36 - 12h + h2 + 36 - 12k + k2

or, 8h + 16k + 8 = 72

or, 8h + 16k = 72 - 8

or, 8(h + 2k) = 64

or, h + 2k = 8.......................................(5)

Similarly,

From equation (2) and (4)

(2 - h)2 + (-2, - k)2=(5 - h)2 + (7 - k)2

or,4 - 4h + h2 + 4 + 4k + k2= 25 - 10h + h2 + 49 - 14k + k2

or, -4h + 4k + 8 = 74 - 10h - 14k

or, -4h + 10h + 4k + 14k = 74 - 8

or, 6h + 18k = 66

or, 6(h + 3k) = 66

or, h + 3k = 11......................................(6)

Subtracting equation (6) from equation (5)

h + 2k = 8
h + 3k = 11
- - -
- k = -3

∴ k = 3

Substituting the value of k in equation (6)

h + 3k = 11

or, h + 3× 3 = 11

or, h + 9 = 11

or, h = 11 - 9

∴ h = 2

∴ Centre of circle (h, k) = (2, 3)

Again,

Substituting the value of h, k in equation (2)

(2 - 2)2 + (-2 - 3)2 = a2

or, 0 + (-5)2 = a2

or, a2 = 25

∴ a = 5

Substituting the value of (h, k) and 'a' in equation (1)

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

or, x2 - 4x + 4 + y2 - 6y + 9 = 25

or, x2 + y2 - 4x - 6y + 13 - 25 = 0

∴x2 + y2 - 4x - 6y -12 = 0Ans

Here,

Centre of circle (h, k) and radius of circle = a.

Equation of the circle is: (x - h)2 + (y - k)2 = a2........................(1)

The points (4, 1) and (6, 5) passes through the equation (1)

(4 - h)2 + (1 - k)2 = a2.....................(2)

(6 - h)2 + (5 - k)2 = a2.....................(3)

From equation (2) and (3)

(6 - h)2 + (5 - k)2=(4 - h)2 + (1 - k)2

or, 36 - 12h + h2 + 25 - 10k + k2 = 16 - 8h + h2 + 1 - 2k + k2

or, 4h + 8k = 44

or, 4(h + 2k) = 44

or, h + 2k = 11..........................(4)

Now,

Centre (h, k) lies on the line 4x + y = 16

4h + k = 16

k = 16 - 4h...........................(5)

Substituting the value of k in equation (4)

h + 2(16 - 4h) = 11

or, h + 32 - 8h = 11

or, -7h = 11 - 32

or, -7h = -21

or, h = \(\frac {-21}{-7}\)

∴ h = 3

Substituting the value of h in equation (5)

∴ k = 16 - 4× 3 = 16 - 12 = 4

Substituting (h, k) in equation (2)

(4 - 3)2 + (1 - 4)2 = a2

or, a2= 12+ (-3)2

or, a2 = 1 + 9

or, a2 = 10

∴ a = \(\sqrt {10}\)

The equation of circle is:

(x - h)2 + (y - k)2 = a2

or, (x - 3)2 + (y - 4)2 = (\(\sqrt {10}\))2

or, x2 - 2.x.3 + 32 + y2 - 2.y.4 + 42 = 10

or, x2 + y2 - 6x - 8y + 9 + 16 = 10

or, x2 + y2 - 6x + 8y + 25 - 10 = 0

∴x2 + y2 - 6x + 8y + 15 = 0Ans

Here,

x2 + y2 - 4x - 6y + 11 = 0

or, x2 - 4x + y2 - 6y + 11 = 0

or, x2 - 2.x.2 + 22 - 22 + y2 - 2.y.3 + 32 - 32 + 11 = 0

or, (x - 2)2 + (y - 3)2 - 4 - 9 + 11 = 0

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

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

The equation of circle is:

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

Comparing (1) and (2)

(h, k) = (2, 3)

∴ Centre of circle is a mid-point of the diameter of the circle is (2, 3)

Mid-point of circle = (\(\frac {x_1 + x_2}2\), \(\frac {y_1 + y_2}2\))

or, (2, 3) = (\(\frac {3 + x}2\), \(\frac {4 + y}2\))

2 = \(\frac {3 + x}2\)

or, 3 + x = 4

or, x = 4 - 3

∴ x = 1

3 = \(\frac {4 + y}2\)

or, 4 + y = 6

or, y = 6 - 4

∴ y = 2

∴ The other point is: (1, 2)Ans

Here,

2x2 + 2y2 - 6x - 2y - 13 = 0

or, x2 + y2 - 3x - y - \(\frac {13}2\) = 0 [\(\because\) Dividing by 2 on both sides]

or, x2 - 3x + y2 - y - \(\frac {13}2\)

or, x2 - 2.x.\(\frac 32\) + (\(\frac 32\)2 - (\(\frac 32\)2+ y2- 2.y.\(\frac 12\) + (\(\frac 12\))2 - (\(\frac 12\))2 - \(\frac {13}2\) = 0

or, (x - \(\frac 32\))2 + (y - \(\frac 12\))2 = \(\frac 94\) + \(\frac 14\) + \(\frac {13}2\)

or, (x - \(\frac 32\))2 + (y - \(\frac 12\))2 = \(\frac {9 + 1 + 26}4\)

or, (x - \(\frac 32\))2 + (y - \(\frac 12\))2 = \(\frac {36}4\)

or, (x - \(\frac 32\))2 + (y - \(\frac 12\))2 = (\(\frac 62\))2...............................(1)

The equation of circle is:

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

Comparing equation (1) and (2)

(h, k) = (\(\frac 32\), \(\frac 12\))

radius (a) = \(\frac 62\) units

Length of the diameter = 2r = 2× \(\frac 62\) = 6 units

∴ Centre =(\(\frac 32\), \(\frac 12\)) and diameter = 6 unitsAns

Given equation is:

x2 + y2 - 2x - 6y + 1 = 0

or, x2 - 2x + 1 + y2 - 6y = 0

or, x2 - 2.x.1 + 12 + y2 - 2.y.3 + 32 - 32 = 0

or, (x - 1)2 + (y - 3)2 - 9 = 0

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

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

Comparing equation (1) and (2)

∴ Centre of circle (h, k) = (1, 3)

and radius (r) = 3 unitsAns

Given equation is:

x2 + y2 - 2x - 2y = 8

or, x2- 2.x.1 + 12 - 12 + y2- 2.y.1 + 12 - 12 = 8

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

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

or, (x - 2)2 + (y - 2)2= \(\sqrt {10}\)..........................(1)

The eqnof circle is:

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

Comparing equation (1) and (2)

h = 1

k = 1

r = \(\sqrt {10}\) units

(h, k) = (1, 1)

From the above figure,

mid-point of AB = (\(\frac {x_1 + x_2}2\), \(\frac {y_1 + y_2}2\))

or, (1, 1) = (\(\frac {x_1 + 2}2\), \(\frac {y_1 + 4}2\))

1 = \(\frac {x + 2}2\)

or, x + 2 = 2

or, x = 2 - 2

∴ x = 0

1 = \(\frac {y + 4}2\)

or, y + 4 = 2

or, y = 2 - 4

∴ y = -2

∴ The other end of the diameter is: (0, -2)Ans

Here,

Given equation is:

2x2 + 2y2 - 8x - 12y + 1 = 0

or, \(\frac {2x^2}2\) + \(\frac {2y^2}2\) - \(\frac {8x}2\) - \(\frac {12y}2\) + \(\frac 12\) = 0 [\(\because\) dividing by 2 on both sides]

or, x2 + y2 - 4x - 6y + \(\frac 12\) = 0

or, x2- 4x + y2- 6y + \(\frac 12\) = 0

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

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

or, (x - 2)2 + (y - 3)2 = 13 - \(\frac 12\)

or, (x - 2)2 + (y - 3)2 =\(\frac {26 - 1}2\)

or, (x - 2)2 + (y - 3)2 =\(\frac {25}2\)

or, (x - 2)2 + (y - 3)2 =(\(\frac 5{\sqrt 2})\)2........................(1)

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

Comparing equation (1) and (2)

∴Centre of circle = (h, k) = (2, 3)

and Radius of circle (r) =(\(\frac 5{\sqrt 2})\)Ans

c

Let: centre of circle be (h. k).

Distance of OA = Distance of OB

or, \(\sqrt {(h - 0)^2 + (k + 2)^2}\) = \(\sqrt {(h + 2)^2 + (k - 0)^2}\)

Squaring on both sides,

h2 + k2 + 4k + 4 = h2 + 4h + 4 + k2

or, 4k = 4h

or, k - h = 0.........................(1)

The centre (h, k) lies on the line:

2x - 3y + 1 = 0

2h - 3k + 1 = 0....................(2)

Multiply eqn (1) by 2 and adding with eqn (2)

-2h + 2k = 0
2h - 3k = -1
- k = -1

∴ k = 1

Putting the value of k in eqn (1)

k - h = 0

or, 1 - h = 0

or, -h = -1

∴ h = 1

∴ Centre (h, k) = (1, 1)

\begin{align*} Radius (r) &= \sqrt {(x - h)^2 + (y - k)^2}\\ &= \sqrt {(0 - 1)^2 + (-2 - 1)^2}\\ &= \sqrt {1 + 9}\\ &=\sqrt {10} units\\ \end{align*}

Equation of circle is:

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

or, (x - 1)2 + (y - 1)2 = (\(\sqrt {10}\))2

or, x2 - 2.x1 + 12 + y2 - 2.y.1 + 12 = 10

or, x2 + y2 - 2x - 2y + 1 + 1 - 10= 0

∴x2 + y2 - 2x - 2y - 8 = 0Ans

f

Let: the centre of circle = (h, k)

The centre of a circle having radius 5 units (h, k) lies on the line 2x + y - 1 = 0.

2h + k - 1 = 0

or, k = 1 - 2h.......................(1)

Eqn of circle is:

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

(x - h)2 + (y - k)2 = 52..................(2)

The point (4, 3) passes through the eqn (2)

(4 - h)2 + (3 - k)2 = 25

or, 16 - 8h + h2 + 9 - 6k + k2 = 25

or, h2 - 8h - 6k + k2 + 25 - 25 = 0

or,h2 - 8h - 6k + k2= 0.............................(3)

Putting the value of k in eqn (3)

h2 - 8h - 6(1 - 2h) + (1 - 2h)2 = 0

or, h2 - 8h - 6 + 12h + 1 - 4h + 4h2 = 0

or, 5h2 = 5

or, h2 = \(\frac 55\)

or, h2 = 1

∴ h = ± 1

Putting the value of k = 1 in eqn (1)

k = 1 - 2h = 1 - 2× 1 = -1

∴ (h, k) = (1, -1)

Putting the value of (h, k) in eqn (2)

(x - h)2 + (y - k)2 = 25

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

or, x2 - 2.x.1 + 12 + y2 - 2.y.1 + 12 = 25

or, x2 + y2 - 2x - 2y + 1 + 1 = 25

or,x2 + y2 - 2x - 2y = 25 - 2

∴x2 + y2 - 2x - 2y = 23

Putting the value of h = -1 in eqn (1)

k = 1 - 2h = 1 - 2 × (-1) = 1 + 2 = 3

∴ (h, k) = (-1, 3)

Putting the value of (h, k) in eqn (2)

(x - h)2 + (y - k)2 = 25

or, (x + 1)2 + (y - 3)2 = 25

or, x2 + 2x + 1 + y2 - 6y + 9 = 25

or, x2+ y2 + 2x - 6y + 10 = 25

or, x2+ y2 + 2x - 6y = 25 - 10

∴x2+ y2 + 2x - 6y =15

∴ Required equations are:x2 + y2 - 2x - 2y = 23 andx2+ y2 + 2x - 6y =15Ans

Given equation is:

2x - 6y - x2 - y2 = 1

or, -x2 + 2x - y2 - 6y - 1 = 0

or, -(x2 - 2x + y2 + 6y + 1) = 0

or,x2 - 2x + 1 + y2 + 6y = 0

or, x2 - 2.x.1 + 12 + y2 + 2.y.3 + 32 - 32 = 0

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

Equation of circle when the centre of circle is (h, k) and radius (r) is:

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

Comparing (1) and (2)

∴ Center of circle (h, k) = (1, -3)

and Radius of circle (r) = 3 unitsAns

Here,

2x2 + 2y2 - 2x + 6y = 45

or, \(\frac {2x^2}2\) + \(\frac {2y^2}2\) - \(\frac {2x}2\) + \(\frac {6y}2\) = \(\frac {45}2\) [\(\because\) Dividing by 2 on both sides.]

or, x2 + y2 - x + 3y = \(\frac {45}2\)

or, x2- x + y2 + 3y = \(\frac {45}2\)

or, x2 - 2.x.\(\frac 12\) + (\(\frac 12)^2\) - (\(\frac 12)^2\) + y2 + 2.y.\(\frac 32\) + (\(\frac 32)^2\) - (\(\frac 32)^2\) = \(\frac {45}2\)

or, (x - \(\frac 12\))2 + (y + \(\frac 32\))2 -(\(\frac 14)\) -(\(\frac 94)\) = \(\frac {45}2\)

or, (x - \(\frac 12\))2 + (y + \(\frac 32\))2= \(\frac {45}2\) + \(\frac 14\) + \(\frac 94\)

or, (x - \(\frac 12\))2 + (y + \(\frac 32\))2= \(\frac {90 + 1 + 9}4\)

or, (x - \(\frac 12\))2 + (y + \(\frac 32\))2 = \(\frac {100}4\)

or, (x - \(\frac 12\))2 + (y + \(\frac 32\))2= (\(\frac {10}4)^2\)..........................(1)

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

Comparing equation (1) and (2)

h = \(\frac 12\)

k = \(\frac {-3}2\)

r = \(\frac {10}2\) = 5

∴ Center of the circle (h, k) = (\(\frac 12\), \(\frac {-3}2\))

and Radius (r) = 5 unitsAns

f

Let: C(h, k) be the centre of the circle which passes through P(2, -1), Q(6, 1) and R(6, 3).

CP2 = r2 = (h - 2)2 + (k + 1)2

CQ2 = r2 = (h - 6)2 + (k - 1)2

CR2 = r2 = (h - 6)2 + (k - 3)2

Now,

CP2= CQ2

or,(h - 2)2 + (k + 1)2 =(h - 6)2 + (k - 1)2

or, h2- 4h + 4 + k2 + 2k + 1 = h2 - 12h + 36 + k2 - 2k + 1

or, 8h + 4k - 32 = 0

or, 4(2h + k - 8) = 0

or, 2h + k - 8 = 0...........................(1)

Also,

CP2 = CR2

or, (h - 2)2 + (k + 1)2 = (h - 6)2 + (k - 3)2

or, h2 - 4h + 4 + k2 + 2k + 1 = h2 - 12h + 36 + k2 - 6k + 9

or, 8h + 8k - 40 = 0

or, 8(h + k - 5) = 0

or, h + k - 5 = 0..........................(2)

Subtracting eqn (1) from (2): we get;

h + k - 5 = 0
2h + k - 8 = 0
- - +
-h + 3 = 0

or, -h = -3

∴ h = 3

Putting the value of h in eqn (2)

h + k - 5 = 0

or, 3 + k - 5 = 0

or, k - 2 = 0

∴ k = 2

∴ Centre = (h, k) = (3, 2)

and Radius (r) = \(\sqrt {(2 - 3)^2 + (-1 - 2)^2}\) = \(\sqrt {(-1)^2 + (-3)^2}\) = \(\sqrt {10}\)

Putting the value of (h, k) and r in eqn:

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

or, (x - 3)2 + (y - 2)2 = (\(\sqrt {10}\))2

or, x2 - 6x + 9 + y2 - 4y + 4 = 10

or, x2 + y2 - 6x - 4y + 13 = 10

or, x2 + y2 - 6x - 4y + 13 -10 = 0

∴ x2 + y2 - 6x - 4y +3 = 0Ans

Given eqn is:

x + y = 2

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

x2 + y2 = 4........................(2)

Putting the value of y in equation (2)

x2 + (2 - x)2 = 4

or, x2 + 4 - 4x + x2 = 4

or, 2x2 - 4x + 4 - 4 = 0

or, 2x2 - 4x = 0

or, 2x (x - 2) = 0

Either: 2x = 0 i.e. x = 0

Or: x - 2 = 0 i.e. x = 2

Putting the value of x in eqn (1)

If: x = 0

y = 2 - 0 =2

If: x = 2

y = 2 - 2 = 0

∴ The required points are: (0, 2) and (2, 0)Ans

Given equation of diameters of the circle are:

x + y = 14..........................(1)

2x - y = 4...........................(2)

Adding equation (1) and (2)

x + y = 14
2x - y = 4
3x = 18

or, x = \(\frac {18}3\)

∴ x = 6

Putting the value of x in equation (1)

x + y = 14

or, 6 + y = 14

or, y = 14 - 6

∴ y = 8

Intersection point of the two diameter equation of the circle is a center of a circle (h, k) = (6, 8)

The given circle passes through the point (3, 4)

\begin{align*} Radius\; of\; circle &= \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}\\ &=\sqrt {(6 - 3)^2 + (8 - 4)^2}\\ &= \sqrt {3^2 + 4^2}\\ &= \sqrt {9 + 16}\\ &= \sqrt {25}\\ &= 5\; units\\ \end{align*}

Equation of the circle whose center is (h, k). Then:

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

or, (x - 6)2 + (y - 8)2 = 52

or, x2 - 12x + 36 + y2- 16y + 64 = 25

or, x2 + y2 - 12x - 16y + 100 - 25 = 0

∴ x2 + y2 - 12x - 16y + 75 = 0Ans

Given equations of two diameters are:

2x - y = 5

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

x - 3y + 5 = 0...............(2)

Putting the value of y in eqn (2)

x - 3 (2x - 5) + 5 = 0

or, x - 6x + 15 + 5 = 0

or, -5x = - 20

or, x = \(\frac {-20}{-5}\)

∴ x = 4

Putting the value of x in eqn (1)

y = 2x - 5

or, y = 2× 4 - 5

or, y = 8 - 5

∴ y = 3

The center of the circle (h, k) = (4, 3)

Radius of the circle (r) = 5 units

Equation of the circle is:

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

or, (x - 4)2 + (y - 3)2 = 52

or, x2 - 8x + 16 + y2 - 6y + 9 = 25

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

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

∴ x2 + y2 - 8x - 6y= 0.............................(3)

The point (0, 0) passes through the eqn (3)

02 + 02 - 4× 0 - 6× 0 = 0

0 = 0

∴ The equation x2 + y2 - 4x - 6y = 0 passes through origin.Proved

Given points are: (1, 2) and (3, 6)

Equation of the diameter of a circle is:

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

or, (x - 1) (x - 3) + (y - 2) (y - 6) = 0

or, x2- x - 3x + 3 + y2 - 2y - 6y + 12 = 0

∴ x2 + y2 - 4x - 8y + 15 = 0

Hence, the required equation is:x2 + y2 - 4x - 8y + 15 = 0Ans

Equation of the circle with center at (h, k) and radius (r) is:

(x - h)2 + (y - k)2 = r2.........................(1)

Points A(-4, -2), B(2, 6) and C(2, -2) passes through the equation (1)

(-4 - h)2 + (-2 - k)2 = r2.......................(2)

(2 - h)2 + (6 - k)2 = r2..........................(3)

(2 - h)2 + (-2 - k)2 = r2........................(4)

Taking equation (2) and (3) and solving:

(-4 - h)2 + (-2 - k)2 =(2 - h)2 + (6 - k)2

or, 16 + 8h + h2 + 4 + 4k + k2 = 4 - 4h + h2 + 36 - 12k + k2

or, 8h + 4k + 20 = 40 - 4h - 12k

or, 8h + 4h + 4k + 12k = 40 - 20

or, 12h + 16k = 20

or, 4(3h + 4k) = 20

or, 3h + 4k = \(\frac {20}4\)

or, 3h + 4k = 5........................(5)

Taking equation (3) and (4) and solving:

(2 - h)2 + (6 - k)2 =(2 - h)2 + (-2 - k)2

or, 4 - 4h + h2 + 36 - 12k + k2 = 4 - 4h + h2 + 4 + 4k + k2

or, 40 - 4h - 12k = 8 - 4h + 4k

or, -12k - 4k = 8 - 40

or, -16k = - 32

or, k = \(\frac {-32}{-16}\)

∴ k = 2

Putting the value of k in equation (5)

3h + 4× 2 = 5

or, 3h + 8 = 5

or, 3h = 5 - 8

or, h = \(\frac {-3}3\)

∴ h = -1

Now,

Center (h, k) = (-1, 2)

One point = B(2, 6)

\begin{align*} Radius\; (r) &= \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}\\ &= \sqrt {(2 + 1)^2 + (6 - 2)^2}\\ &= \sqrt {3^2 + 4^2}\\ &= \sqrt {9 + 16}\\ &= \sqrt {25}\\ &= 5\; units\\ \end{align*}

∴ Radius of circle is 5 units.Ans

Let: r be the radius of the circle.

Then: its equation is:

(x - 2)2 + (y - 1)2 = r2..........................(1)

The given line will be tangent to the circle. If it's perpendicular distance from the center is equal to the radius.

\begin{align*} r &= \begin {vmatrix} \frac {Ax\;+\; By\;+\; C}{\sqrt {A^2 + B^2}} \end {vmatrix}\\ &= \begin {vmatrix} \frac {3 × 2 + (-4) × 1 + 1}{\sqrt {(3)^2 + (-4)^2}}\\ \end {vmatrix}\\ &= \begin {vmatrix} \frac {6-4+1}{\sqrt {9 + 16}}\\ \end {vmatrix}\\ &= \begin {vmatrix} \frac {3}{\sqrt {5}}\\ \end {vmatrix}\\ &= \frac 35\\ \end{align*}

Now,

Putting the value of r in equation (1)

(x - 2)2 + (y - 1)2 = (\(\frac 35)^2\)

or, x2 - 4x + 4 + y2 - 2y + 1 = \(\frac 9{25}\)

or, 25x2 - 100x + 100 + 25y2 - 50y + 25 - 9 = 0

or, 25x2 + 25y2 - 100x - 50y + 116 = 0

Hence, the required equation is:25x2 + 25y2 - 100x - 50y + 116 = 0Ans

Here,

The eqn of circle is:

x2 + y2 - 2x + 6y - 39 = 0

or, x2 - 2.x.1 + 12 - 12 + y2 + 2.y.3 + 32 - 32 - 39 = 0

or, (x - 1)2 + (y + 3)2 - 1 - 9 - 39 = 0

or, (x - 1)2 + (y + 3)2 - 49 = 0

or, (x - 1)2 + (y + 3)2= 49

or, (x - 1)2 + (y + 3)2=72........................(1)

The eqn of circle is:

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

Comparing equation (1) and (2)

Center of circle (h, k) = (1, -3)

Radius (r) = 7

The given equation of a line is:3x + y + 7\(\sqrt {10}\) = 0

The length of the perpendicular drawn from the center (1, -3) on the line3x + y + 7\(\sqrt {10}\) = 0 is:

\begin{align*} d &= \frac {3 × 1 - 3 + 7\sqrt {10}}{\sqrt {3^2 + 1^2}}\\ &= \frac {7\sqrt {10}}{\sqrt {10}}\\ &= 7\\ \end{align*}

Length of the perpendicular = radius of the circle = 7 units

∴ 3x + y + 7\(\sqrt {10}\) = 0 is tangent to the circle.Proved

Here,

The equation of circle is:

x2 + y2 + 6x - 2y - 15 = 0

or, x2 + 2.x.3 + 32 - 32 + y2 - 2.y.1 + 12 - 12 - 15 = 0

or, (x + 2)2 + (y - 1)2 - 9 - 1 - 15 = 0

or, (x + 2)2 + (y - 1)2 - 25 = 0

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

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

The eqn of circle is:

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

Comparing (1) and (2)

Center (h, k) = (-3, 1) and radius (r) = 5

The point of contact is: A(2, 2)

Slope of OA = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {2 -1}{2 + 3}\) = \(\frac 15\)

Slope of the tangent PQ (m) = -5

Equation of the tangent PQ at (2, 2) is:

y - y1 = m(x - x1)

or, y - 2 = -5(x - 1)

or, y - 2 = -5x + 5

or, 5x + y = 5 + 2

∴ 5x + y = 7Ans

Let: The center of circle is be (h, k) and radius be r.

Equation of circle is:

(x - h)2 + (y - h)2 = r2....................(1)

The point (3, 2) and (5, 4) passes through eqn (1)

(3 - h)2 + (2 - k)2 = r2....................(2)

(3 - h)2 + (4 - k)2 = r2....................(3)

From (2) and (3)

(3 - h)2 + (2 - k)2 =(3 - h)2 + (4 - k)2

or, 9 - 6h + h2 + 4 - 4k + k2 = 25 - 10h + h2 + 16 - 8k + k2

or, 4h + 4k - 28 = 0

or, 4(h + k - 7) = 0

or, h + k - 7 = 0

or, h = 7 - k.................(4)

The point (h, k) lines in the line:

3x - 2y = 1

3h - 2k = 1...................(5)

Putting the value of (h, k) in eqn(5)

3(7 - k) - 2k = 1

or, 21 - 3k - 2k = 1

or, 21 - 5k = 1

or, -5k = 1 - 21

or, -5k = -20

or, k = \(\frac {-20}{-5}\)

∴ k = 4

Putting the value of k in eqn (4)

h = 7 - k = 7 - 4 = 3

Putting the value of (h, k) in eqn (2)

(3 - 3)2 + (2 - 4)2 = r2

or, 0 + (-2)2 = r2

or, r2 = 4

∴ r = 2

Putting the value of (h, k) in eqn (1)

(x - 3)2 + (y - 4)2 = 22

or, x2 - 6x + 9 + y2 - 8y + 16 = 4

or, x2 + y2 - 6x - 8y + 25 - 4 = 0

∴ x2 + y2 - 6x - 8y + 21 = 0 Ans

Here,

x + 2y - 1 = 0.....................(1)

2x - y - 7 = 0......................(2)

Eqn (1) is multiplied by 2 and adding with eqn (1)

4x - 2y - 14 = 0
x + 2y - 1 = 0
5x - 15 = 0

or, 5x = 15

or, x = \(\frac {15}5\)

∴ x = 3

Putting the value of x in eqn (1)

x + 2y - 1 = 0

or, 3 + 2y - 1 = 0

or, 2y = -2

or, y = -\(\frac 22\)

∴ y = -1

The center of circle = (h, k) = (3, -1)

\begin{align*} The\; radius\; of\; the\; circle\; (r) &= \sqrt {(3 - 3)^2 + (1 - (-1))^2}\\ &= \sqrt {0 + (-2)^2}\\ &= \sqrt 4\\ &= 2\; units \end{align*}

The eqn of circle is: (x - h)2 + (y - k)2 = r2.........................(3)

Putting the value of (h, k) and r in eqn (2)

(x - 3)2 + (y + 1)2 = 22

or, x2- 6x + 9 + y2+ 2y + 1 = 4

or, x2 + y2 - 6x + 2y + 10 - 4 = 0

∴ x2 + y2 - 6x + 2y + 6 = 0Ans

Let:

Center of the circle be (h, k) and radius be r.

Equation of the circle is:

(x - h)2 + (y - k)2 = r2......................(1)

The point (0, 0) and (4, 2) are passing through the eqn (1)

(0 - h)2 + (0 - k)2 = r2

h2 + k2 = r2.................(2)

(4 - h)2 + (2 - k)2 = r2.......................(3)

From eqn (2) and (3)

h2 + k2 =(4 - h)2 + (2 - k)2

or, h2 + k2 = 16 - 8h + h2 + 4 - 4k + k2

or, 8h + 4k - 20 = 0

or, 4(2h + k - 5) = 0

or, 2h + k - 5 = 0

or, k = 5 - 2h...........................(4)

The point (h, k) lies on the line:

x + y = 1

h + k = 1........................(5)

Putting the value of k in eqn (5)

h + 5 - 2h = 1

or, 5 - 1 = h

∴h = 4

Putting the value of h in eqn (4)

k = 5 - 2h = 5 - 2× 4 = 5 - 8 = -3

Putting the value of (h, k) = (4, -3) in eqn (2)

h2 + k2 = r2

or, r2 = 42 + (-3)2

or, r2 = 16 + 9

or, r2 = 25

∴ r = 5

Putting the value of (h, k) and r in the eqn

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

or, (x - 4)2 + (y + 3)2 = 52

or, x2- 8x + 16 + y2 + 6y + 9 = 25

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

∴x2 + y2 - 8x + 6y = 0Ans

0%
  • x+ y- 20y + 75 = 0

    (0 , 10)


    (0 , 0)


    (10 , 20)


    (10 , 20)


  • x2 + y2 + 4x - 5 = 0 

    (0 , 2)


    (0)


    (-2 , 0)


    (2 , 0)


  • x2 + y2 - 6x + 2y + 1  = 0

    (3 , -1)


    (-1 , -2)


    (-1 , -1)


    (2 , -1)


  • Find the length and radius of the circle . 


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

    24 units


    5 units


    16 units


    3 units


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

    44 units


    24 units


    96 units


    98 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 + 4x - 2y = -4


    x2 + y2 + 2x + 4y = 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 - 2x + 4y = 0


    x2 - y2 + 2x + 4y = 0


    x2 - y+ 2x - 4y = 0


    x2 - y2 - 4x + 2y = 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) ?

    16 units


    3 units


    2 units


    4 units


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

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


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


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


    x2 - y2 + 6x - 12 + 9y = 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 - 15  = 0


    x2 - y2 + 4x + 8y - 15  = 0


    x2 - y2 + 4x + 8y + 0  = 15


  • 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 + 4y + 0 = 7


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


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

    x2 + y2 - 2x - 6y +  1 = 0

    4 , (1 , 3) units


    (1 , 3)  , 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) , 3 units


    (3 , 1)  units


    (-1 , -3) , 1 units


  • You scored /15


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Any Questions on Circle ?

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Discussions about this note

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Bikesh Shreatha

Find the equation of a circle with center at (-3,2) and passing through the centre of a circle x^2 y^3-2x-10y-20=0

find the equation of the circle having radius 6and touching both exes

find the equation of the circle having radius 6 and touching both exes

api

A circle passes through origin and makes the intercepts 6 units and 8 units on x-axis and y-axis respectively.find its centre,diameter and equation.