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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.
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,
OP^{2}= r^{2}
or, (x - 0)^{2} + (y - 0)^{2} = r^{2}
or, x^{2} + y^{2} = r^{2}
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,
QP^{2} = r^{2}
or, (x - h)^{2} + (y k)^{2} = r^{2}
This is the equation of the circle.
3. Circle with a given diameter (Diameter form)
Let A (x_{1}, y_{1}) and B(x_{2}, y_{2}) 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 - y_{1}) . (y - y_{2}) = - (x - x_{1}) . (x - x_{2})
or, (x - x_{1}) . (x - x_{2}) +(y - y_{1}) . (y - y_{2}) = 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,
QP^{2} = r^{2}
or, (x - h)^{2} + (y - k)^{2} = r^{2}
or, x^{2} - 2hx + h^{2} + y^{2} - 2ky + k^{2} = r^{2}
or, x^{2} + y^{2} - 2hx - 2ky + h^{2} + k^{2} - r^{2} = 0
This is the equation of a circle having centre at the point (h, k) and radius r.
Putting -2h = 2g, -2k = 2f and h^{2} + k^{2} - r^{2} = c we have,
x^{2} + y^{2} + 2gx + 2fy + c = 0
This is general equation of the circle.
Note: The general equation of a circle is: x^{2} + y^{2} + 2gx + 2fy + c = 0.
Comparing this equation with the general equation of second degree
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 we have,
- coefficient of x^{2} and y^{2} are equal
- coefficient of xy is zero
Hence,
The general equation of second degreeax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a circle if
- coefficients of x^{2} and y^{2} are equal i.e. a = b
- coefficient of xy is zero i.e. h = 0
The general equation of a circle is:
x^{2} + y^{2} + 2gx + 2fy + c = 0
or, x^{2} + 2gx + y^{2}+ 2fy = -c
or, x^{2} + 2gx + g^{2} + y^{2} + 2fy + f^{2} = g^{2} + f^{2} - 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} = r^{2}, 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) = \(\sqrt {g^2 + f^2 - c}\)
Let, A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) be three points of a circle.
Let, P (h, k) be the centre of the circle.
Then,
AP = BP = CP
or, AP^{2} = BP^{2} = CP^{2}
By using distance formula,
We have,
AP^{2}= (x_{1} - h)^{2} + (y_{1} - k)^{2}
BP^{2}= (x_{2} - h)^{2} + (y_{2} - k)^{2}
CP^{2}= (x_{3} - h)^{2} + (y_{3} - k)^{2}
Taking AP^{2} = BP^{2} we have,
(x_{1} - h)^{2} + (y_{1}- k)^{2} = (x_{2} - h)^{2} + (y_{2} - k)^{2} .....................................................(i)
Taking BP^{2} = CP^{2} we have,
(x_{2} - h)^{2} + (y_{2}- k)^{2} = (x_{3} - h)^{2} + (y_{3} - 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} = r^{2} we get the required equation of the circle.
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} = r^{2}
or, (x - h)^{2 }+ (y - k)^{2}= k^{2}
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} = r^{2}
or,(x - h)^{2} + (y - k)^{2} = h^{2}
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} = r^{2}
or,(x - h)^{2} + (y - k)^{2} = h^{2}
or,(x - k)^{2} + (y - h)^{2} = k^{2}
or,(x - r)^{2} + (y - k)^{2} = k^{2}
Equation of a circle
Equation of a circle in particular cases
Here,
2x^{2} + 2y^{2} - 5x - 7y - 23 = 0
or, \(\frac {2x^2}2\) + \(\frac {2y^2}2\) - \(\frac 52\)x - \(\frac 72\)y - \(\frac {23}2\) = 0
or, x^{2} - \(\frac 52\)x + y^{2} - \(\frac 72\)y - \(\frac {23}2\) = 0
or, x^{2} - 2.x.\(\frac 54\) + (\(\frac 54)\)^{2} + y^{2} - 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)
Eq^{n} of circle is: (x - h)^{2}+ (y - x)^{2} = r^{2}......................(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,
x^{2} + y^{2} - 20y + 75 = 0
or, x^{2} + y^{2} - 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 eq^{n} of circle is: (x - h)^{2} + (y - k)^{2} = r^{2}.....................(2)
Comparing (1) and (2)
Centre of circle (h, k) = (0. 10) _{Ans}
Here,
x^{2} + y^{2} - 4x + 10y - 7 = 0
or, x^{2}- 4x + y^{2} + 10y - 7 = 0
or, x^{2} - 2.x.2 + 2^{2} - 2^{2} + y^{2} + 2.y.5 + 5^{2} - 5^{2} - 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}= 6^{2}...........................(1)
The eq^{n} of the circle is: (x - h)^{2}+ (y - k)^{2} = r^{2}.........................(2)
Comparing (1) and (2)
The length of radius of a circle is: 6 units _{Ans}
Calculate the length of the radius of the circle whose equation is: x^{2} + y^{2} + 4x - 6y + 4 = 0.
Here,
x^{2} + y^{2} + 4x - 6y + 4 = 0
or, x^{2} + 4x+ 4 + y^{2} - 6y = 0
or, x^{2} + 2.x.2 + 2^{2}+ y^{2} - 2.y.3 + 3^{2} - 3^{2}= 0
or, (x + 2)^{2 }+ (y - 3)^{2} = 3^{2}.....................................(1)
The eq^{n} of circle is: (x - h)^{2} + (y - k)^{2} = r^{2}...........................(2)
Comparing (1) and (2)
The length of radius of the circle (r) = 3 units_{Ans}
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} = r^{2}
(x - 3)^{2} + (y - 0)^{2} = 5^{2}
or, x^{2} - 6x + 9 + y^{2} = 25
or, x^{2} + y^{2} - 6x + 9 - 25 = 0
∴x^{2} + y^{2} - 6x -16 = 0_{Ans}
Here,
x^{2} + y^{2} + 4x - 4y -1 = 0
or, x^{2} + 4x + y^{2} - 4y - 1 = 0
or, x^{2} + 2.x.2 + 2^{2} - 2^{2} + y^{2} - 2.y.2 + 2^{2} - 2^{2} - 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}=3^{2}................................(1)
Equation of circle, (x - h)^{2} + (y - k)^{2} = r^{2}...................................(2)
Comparing (1) and (2)
h = -2
k = 2
r = 3
∴ Centre = (-2, 2) and radius = 3 units_{Ans}
Here,
x^{2}+ y^{2} - 2y = 24
or, x^{2} + y^{2} - 2.y.1 + 1^{2}- 1^{2} = 24
or, x^{2} + (y - 1)^{2} - 1 = 24
or, x^{2} + (y - 1)^{2} = 25
or, x^{2} + (y - 1)^{2} = 5^{2}.............................(1)
The equation of circle is: (x - h)^{2} + (y - k)^{2} = a^{2}........................(2)
Comparing (1) and (2)
(h, k) = (0, 1)
a = 5
Hence, centre = (0, 1) and radius = 5 units_{Ans}
Given eq^{n} is:
(x + 5)^{2} + y^{2} = 121
or, (x + 5)^{2} + (y - 0)^{2} = (11)^{2}..........................(1)
The eq^{n} of the circle is:
(x - h)^{2} + (y - k)^{2} = r^{2}..............................(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 units_{Ans}
Here,
x^{2} + y^{2} - 4x - 6y - 12 = 0
or, x^{2} - 4x + y^{2} - 6y - 12 = 0
or, x^{2} - 2.x.2 + 2^{2} - 2^{2} + y^{2}- 2.y.3 + 3^{2}- 3^{2}- 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} = 5^{2}............................(1)
The equation of circle is: (x - h)^{2} + (y - k)^{2} = r^{2}.............................(2)
Comparing (1) and (2)
(h, k) = (3, 2) and r = 5
∴ The centre of circle = (3, 2) and radius of circle (r) = 5 units_{Ans}
Here,
(x_{1} - y_{1}) = (2, 4)
(x_{2} - y_{2}) = (3, -6)
The equation of a circle in diameter form:
(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0
or, (x - 2) (x - 3) + (y - 4) (y + 6) = 0
or, x^{2} - 3x - 2x + 6 + y^{2} + 6y - 4y - 24 = 0
or, x^{2} - 5x + 6 + y^{2} + 2y - 24 = 0
∴ x^{2} + y^{2} - 5x + 2y - 18 = 0_{Ans}
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 - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0
or, (x, 4) (x, 3) + (y - 2) (y - 5) = 0
or, x^{2} - 3x - 4x + 12 + y^{2} - 5y - 2y + 10 = 0
∴ x^{2} + y^{2} - 7x - 7y + 22 = 0_{Ans}
Here,
Radius of circle (r) = 5 units
Eq^{n} of two diameters are:
x = 3y.............................(1)
y = 2...............................(2)
Putting the value of y in eq^{n} (1)
x = 3× 2 = 6
Intersection point of the two diameters is centre of circle so:
Centre of circle = (h, k) = (6, 2)
Eq^{n} of circle is:
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 6)^{2} + (y - 2)^{2} = 5^{2}
or, x^{2}- 12x + 36 + y^{2} - 4y + 4 - 25 = 0
∴ x^{2} + y^{2} - 12x - 4y + 15 = 0_{Ans}
Here,
Centre of circle (h, k) = (-2, 3)
Radius (r) = k = 3
The eq^{n} of circle is:
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x + 2)^{2} + (y - 3)^{2} = 3^{2}
or, x^{2} + 4x + 4 + y^{2} - 6y + 9 = 9
or, x^{2} + y^{2}+ 4x - 6y + 13 - 9 = 0
∴ x^{2} + y^{2}+ 4x - 6y + 4 = 0_{Ans}
Here,
Centre of circle (h, k) = (4, -3)
Radius (r) = h = 4
The equation of circle is:
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 4)^{2} + (y + 3)^{2} = 4^{2}
or, x^{2} - 8x + 16 + y^{2} + 6y + 9 = 16
or, x^{2} + y^{2} - 8x + 6y + 25 = 16
or, x^{2} + y^{2} - 8x + 6y + 25 -16 = 0
∴ x^{2} + y^{2} - 8x + 6y + 9 = 0_{Ans}
Here,
Radius of circle (r) = 5 units
h = k = r = 5 [\(\because\) touches on both axis]
The eq^{n} of circle is:
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 5)^{2} + (y - 5)^{2} = 5^{2}
or, x^{2} - 10x + 25 + y^{2} - 10y + 25 = 25
or, x^{2} + y^{2} - 10x - 10y + 50 - 25 = 0
∴ x^{2} + y^{2} - 10x - 10y + 25 = 0_{Ans}
Here,
Centre of circle = (h, k) = (6, 8)
The eq^{n} of circle is:
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 6)^{2} + (y - 8)^{2} = r^{2}..............................(1)
The eq^{n} (1) passes through the point (0, 0)
(0 - 6)^{2}+ (0 - 8)^{2} = r^{2}
or, 36 + 64 = r^{2}
or, 100 = r^{2}
or, (r)^{2} = (10)^{2}
∴ r = 10 units
Putting the value of r in eq^{n} (1)
(x - 6)^{2} + (y - 8)^{2} = (10)^{2}
or, x^{2}- 12x + 36 + y^{2} - 16y + 64 = 100
or, x^{2} + y^{2} - 12x - 16y + 100 = 100
∴x^{2} + y^{2} - 12x - 16y= 0_{Ans}
x^{2 }+ y^{2 }- 20y + 75 = 0
(10 , 20)
(0 , 10)
(0 , 0)
(10 , 20)
x^{2} + y^{2} + 4x - 5 = 0
(-2 , 0)
(0)
(0 , 2)
(2 , 0)
x^{2} + y^{2} - 6x + 2y + 1 = 0
(-1 , -1)
(2 , -1)
(-1 , -2)
(3 , -1)
Find the length and radius of the circle .
x^{2} + y^{2} + 4x - 6y + 4 = 0
24 units
16 units
3 units
5 units
Find the length of the circumference of a circle having the equation x^{2} + y^{2}- 2y-48 = 0
24 units
98 units
44 units
96 units
Find the equation of the circle having centren (0 , 0) and radius 3 units .
x^{2} - y^{2} = 9
x^{2} - y^{2} = 10
x^{2} + y^{2} = 9
x^{2} + 9 = y^{2}
Find the equation of the circle with centre (2 , -1) and radius 3 units .
x^{2} + y^{2} + 4x + 2y = -4
x^{2} - y^{2} + 4x - 2y = -4
x^{2} + y^{2} - 4x + 2y = 4
x^{2} + y^{2} + 2x + 4y = 4
Find the equation of a circle having centre (1 , -2) and the radius 2 (sqrt{5}) units .
x^{2} - y^{2} - 4x + 2y = 0
x^{2} + y^{2} - 2x + 4y = 0
x^{2} - y^{2} + 2x + 4y = 0
x^{2} - y^{2 }+ 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) ?
3 units
2 units
4 units
16 units
Find the equation of the circle having centre (3 , 6) and touching the x-axis.
x^{2} + y^{2} -6x - 12y + 9 = 0
x^{2} - y^{2} + 6x - 12 + 9y = 0
x^{2} - y^{2} -6x + 11y + 9 = 0
x + y - 6x - 12y + 9 = 0
Given points are the end of the circle . Find the equation of the circle.
A(5 , 6) and B(3 , 4)
x^{2} + y^{2} + 8x + 10y - 39 = 0
x^{2} + y^{2} - 8x - 10y + 39 = 0
x^{2} + y^{2} + 8x - 10y + 0 = 39
x^{2} - y^{2} + 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 - 7 = 0
x2 + y2 -6x + 4y + 0 = 7
x2 - y2 -6x - 6y - 7 = 0
x2 - y2 - 6x -4y - 7 = 0
Find the coordinates of centre and radius of the circle.
x^{2} + y^{2} - 2x - 6y + 1 = 0
1 ,(4 , 3) units
(3 , 1) , 4 units
(1 , 3) , 3 units
4 , (1 , 3) units
Find the coordinates of centre and radius of the circle.
2x - 6y - x^{2} - y^{2 }= 1
(-1 , -3) , 1 units
(1 , 3) , 3 units
(1 , -3) , 3 units
(3 , 1) units
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Jun 14, 2017
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Sagunsharma
Find the equation of circle passing through the point (6,-1) and touch the y axis at (0,5)
Mar 23, 2017
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Rachana
If the centre of circle x^2 y^2 - ax - by - 12 = 0 is (2,3),find the values of a and b.
Mar 15, 2017
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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.
Mar 05, 2017
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