Circle

Engineers exploit the circle's symmetrical properties as seen by the use of
the circle in making watches, clocks, bicycles, cars, trains, ships, aeroplanes,
radios, telephones, trolleys, wheel-barrows, air-conditioners, rockets etc.
Where do we find circle in our real life and how can we use it?

Equation of circle with center at the origin and radius
‘r’ unit:
Let O(0,0) be the center of circle and r be the radius .
Let P(x,y) be any point on the circle and OP= r units.
By using distance formula: OP2 = (x-0)2 +(y-0)2
i.e. r2 =x2 +y2 which is satisfied by each point on the circle.
Hence, it represents the equation of a circle with center at origin
and radius ‘r’ units.

Equation of circle in standard form:
Let A(h, k) be a center of the circle and P(x,y) be any
point on the circle and AP= r units.
By using distance formula: AP2 = (x-h)2 +(y-k)2
Or, r2 =(x-h)2 +(y-k)2
i.e. r2 =(x-h)2 +(y-k)2
Hence, it represents the equation of a circle with
center at A(h,k) and radius ‘r’ units.

Some special cases of circles equations
1. The centre of circle is at origin. x2 + y2 = r2
2. The centre of circle is (h, k), (x – h)2 + (y – k)2 = r2
3. Two ends of diameter are given. (x – x1) (x – x2) + (y – y1) (y – y2) = 0.
4. Passing through origin (r2 = h2 + k2 ).
5. Circle touches X-axis (r = k). (x – h)2 + (y – k)2 = k2 (‡ r = k)
6. Circle touches Y-axis (r = h). (x – h)2 + (y – k)2 = h2 (‡ r = h)
7. Circle touches both the axes (r = h = k).
(x – h)2 + (y – h)2 = h2
or, (x – k)2 + (y – k)2 = k2
or, (x – r)2 + (y – r)2 = r2
(r = h = k)

General equation of circle:
Considering the general equation of a circle is; x2 + y2 + 2gx + 2fy + c = 0
It can be written as, x2 + 2gx + y2 + 2ƒy + c = 0
or, x2 + 2gx + g2+ y2 + 2ƒy + ƒ2 + c = g2 + ƒ 2
or, (x + g) 2 + (y + ƒ)2 = g2 + ƒ2 – c
or, {x – (– g)}2 + {y – (– ƒ)}2 = ( g2 + ƒ2 – c ) 2
Comparing this equation with (x – h) 2 + (y – k) 2 = r2
then, h = – g, k = – ƒ and r = ( g2 + ƒ2 – c )
Thus, centre of the circle (h, k) = (– g, – f) and radius (r) = ( g2 + ƒ2 – c ) .
When general equation is given.
Note:
If g2 + ƒ2 – c >0, the radius is real, hence the equation gives a real geometrical locus.
If g2 + ƒ2 – c =0, the radius is zero, this is a point circle.
If g2 + ƒ2 – c <0; the radius become imaginary. The equation represents a circle with a real centre
and an imaginary radius.

The equation of a circle when coordinates of ends of diameter
are given.
Here, let, AB be a diameter of circle with centre at P where
coordinates of A and B are (x1, y1) and (x2, y2) respectively.
C(x, y) is any point on the circle. Join A, C and B, C.
Now,
slope of AC is m1 =
y – y1
x – x1
and Slope of BC is m2 =
y – y2
x – x2
Here, ACB = 90°, (Angle at semicircle)
So, AC and BC are perpendicular to each other.
Since, AC  BC, m1 × m2 = – 1
⸫Slope of AR × slope of BR = –1
or, (
y – y1
x – x1
) (
y – y2
x – x2
) = -1
or, (y – y1) (y – y2) = – (x – x1) (x – x2)
or, (x – x1) (x – x2) + (y – y1) (y – y2) = 0
Thus, (x – x1 ) (x – x2 ) + (y – y1 ) (y – y2 ) = 0 is the equation of the
circle in diameter form.
C(x,y)
A(x1, y1) B(x2,y2)
P

Find the equation of a circle with center at the origin and
radius 5 units.
Solution,
Given that , O(0,0) be the center of circle and 5units be the radius .
We know r2 =x2 +y2
Or, x2 +y2 = 52
Or, x2 +y2 =25 which is required equation of a circle.

Find the equation of circle having center (2,3) and radius is 6 units.
Solution:
Here given center f a circle (h, k)= (2,3)
And radius (r )= 6 units
We have ,
Equation of circle, r2 =(x-h)2 +(y-k)2
Or, 62 =(x-2)2 +(y-3)2
Or, x2 -4x+4+y2-6y+9=36
Or x2 +y2-4x-6y-23=0 is required equation of circle.

The end points of a diameter of a circle are (3, 0) and (0, 4). Find the
equation of the circle.
Solution:
Here, let (x1, y1) = (3, 0) and (x2, y2) = (0, 4)
We know that,
Equation of circle in diameter form is;
Or, (x – x1) (x – x2) + (y – y1) (y – y2) = 0
Or, (x – 3) (x – 0) + (y – 0) (y – 4) = 0
or, x2 – 3x + y2 – 4y = 0
or, x2 + y2 – 3x – 4y = 0
Thus, the equation of circle is x2 + y2 – 3x – 4y = 0.

Write the equation of the circle having centre at (3, 3)
and passing through (– 2, – 4).
Solution:
Here, Centre C(h, k) = (3, 3) and any point P(x, y) = (– 2, – 4)
Radius (r2) = {(3 + 2)2 + (3 + 4)2 }= 25 + 49 = 74
Now, using (x – h) 2 + (y – k) 2 = r2 formula,
we get, (x – 3)2 + (y – 3)2 = 74
or, x2 – 6x + 9 + y2 – 6y + 9 = 74
or, x2 + y2 – 6x – 6y = 56
Thus, x2 + y2 – 6x – 6y = 56 is the required equation of the circle.

Find the equation of a circle with centre at (2, 3) and passing through the
point (5, 6).
Solution:
Here, centre of circle (h, k) = (2, 3) and passing point (x, y) = (5, 6)
Now, radius of circle r2= (x – h)2 + (y – k)2 = (5 – 2)2 + (6 – 3)2 = 9 + 9 = 18 units
We know that, equation of circle having centre (2, 3) and radius 18 units is;
Or, (x – h)2 + (y – k)2 = r2
Or, (x – 2)2 + (y – 3)2 = 18
Or, x2 + y2 – 4x – 6y + 4 + 9 = 18
Or, x2 + y2 – 4x – 6y – 5 = 0 which is required equation.

Find the equation of the circle having centre (3, 6) and touching the
x-axis.
Solution:
Here, centre = (h, k) = (3, 6)
The circle touches x-axis so k = r = 6
We know that, equation of circle is; (x – h)2 + (y – k)2 = r2
i.e. (x – 3)2 + (y – 6)2 = 62
or, x2 – 6x + 9 + y2 – 12y + 36 = 36
Or, x2 + y2 – 6x – 12y + 9 = 0
Thus, the equation of circle is; x2 + y2 – 6x – 12y + 9 = 0.

Find the equation of the circle having centre at (5, 4) and that touches
the Y-axis.
Solution:
Here, Centre of circle (h, k) = (5, 4)
i.e. h = 5 and k = 4
Given circle touches Y-axis; So,
Radius (r) = h = 5
We know that; the equation of circle; (x – h)2 + (y – k)2 = r2
or, (x – 5)2 + (y – 4)2 = 52
or, x2 – 2.x.5 + 52 + y2 – 2.y.4 + 42 = 25
or, x2 – 10x + 25 + y2 – 8y + 16 = 25
Or, x2 + y2 – 10x – 8y + 16 = 0
Thus, x2 + y2 – 10x – 8y + 16 = 0 is the required equation of circle.

Find the centre and radius of a circle having the
equation x2 + y2 – 10x – 4y = 7.
Solution:
Here, x2 + y2 – 10x – 4y = 7
or, x2 – 10x + y2 – 4y = 7
or, x2 – 2.x.5 + 52 + y2 – 2.y.2 + 22 = 36
or, (x – 5)2 + (y – 2)2 = 62
Comparing it with (x – h)2 + (y – k)2 = r2
Then, centre (h, k) = (5, 2) and radius (r) = 6
Thus, the centre and radius are (5, 2) and 6 units.

If the centre of the circle x2 + y2 – ax – by – 12 = 0 is (2, 3),
find the values of a and b.
Solution:
Here, Given that centre of a circle (h,k)= (2,3) and equation of circle
x2 + y2 – ax – by – 12 = 0 ………………….(A)
Comparing it with General equation of a circle x2+y2+2gx+2fy+c=0
Then we get , 2g= -a i.e g= -a/2
And 2f= -b i.e. f= -b/2
Now centre of the circle (h,k)= (-g.-f)
Or , (2,3) = (a/2, b/2)
Or, a= 4 and b= 6 Ans//

Find the centre and radius of the circle represented by
x2 + (y + 2) 2 = 49.
Solution:
Here, Given equation of a circle , x2 + (y + 2)2 = 49
or, (x – 0)2 + (y + 2)2 = 72
It is in the form of (x – h)2 + (y – k)2 = r2
where, centre (h, k) = (0, – 2) Radius (r) = 7
Thus, (0, – 2) is a centre & 7 units is the radius of given circle

Find the length of the circumference of a circle having
the equation x2 + y2 – 2y – 48 = 0.
Solution:
Here, equation of given circle is; x2 + y2 – 2y – 48 = 0
or, x2 + y2 – 2.y.1 + 12 – 1 – 48 = 0
or, (x – 0)2 + (y – 1)2 = 49
Or, (x – 0)2 + (y – 1)2 = 72
Which is the form of (x – h)2 + (y – k)2 = r2
Where, (h, k) = (0, 1) & r = 7
Now, circumference of circle (c) = 2πr = 2. 22/7 × 7 = 44 units
Thus, circumference of given circle is 44 units.

Find the equation of the circle whose diameter has the
ends at A(– 3, – 4) and B(3, 4).
Solution:
Here, Ends of diameter are A(– 3, – 4) and B(3, 4).
Using formula, (x – x1) (x – x2) + (y – y1) (y – y2) = 0
or, (x + 3) (x – 3) + (y + 4) (y – 4) = 0
or, x2 – 9 + y2 – 16 = 0
Or, x2 + y2 = 25
Thus, x2 + y2 = 25 is the required equation.

Find the centre and radius of the circle 2x2 + 2y2 – 20x – 28y + 98 = 0.
Solution:
Here, given equation of circle is, 2x2 + 2y2 – 20x – 28y + 98 = 0 ………….A
Dividing by 2 to eq(A) then we get,
or, x2 + y2 – 10x – 14y + 49 = 0
or, x2 – 10x + y2 – 14y + 49 = 0
or, x2 – 2 . 5 . x + 25 + y2 – 2 . 7 . y + 49 = 25
or, (x – 5)2 + (y – 7)2 = 52
Comparing this equation with the equation of Circle (x – h)2 + (y – k)2 = r2 ,
we get h = 5, k = 7 and r = 5
Centre (h, k) = (5, 7) and radius = 5 units.
Thus, coordinates of centre is (5, 7) and radius is 5 units.

If the coordinates of one end of a diameter of the circle having equation
x2 + y2 + 4x – 6y + 8 = 0 is (0, 2) then find the co-ordinates of the other
end of the diameter.
Solution:
Here, given equation of circle is; x2 + y2 + 4x – 6y + 8 = 0
or, x2 + 4x + y2 – 6y + 8 = 0
or, x2 + 4x + 4 + y2 – 6y + 9 = 9 + 4 – 8
or, (x + 2)2 + (y – 3)2 = 5
Or, (x + 2)2 + (y – 3)2 = 52
Comparing it with (x – h)2 + (y – k)2 = r2
then, centre (h, k) = (– 2, 3)
By the question, one end of diameter is (0, 2)
Let other end be (x2, y2).
Centre is the midpoint of diameter,
so using formula, x =
x1 + x2
2
and y =
y1 + y2
2
or, -2=
0 + x2
2
And 3=
2 + y2
2
Or, x2= -4 and y2= 4
Thus, the co-ordinates of other end is (– 4, 4).

Find the equation of the circle having centre (1, 2) and passing through
the point of intersection of the lines x + 2y = 3 and 3x + y = 4.
Solution:
Here, Given equations are; x + 2y = 3 … (i)
and 3x + y = 4 …….. (ii)
Solving equation (i) and (ii) then, (x + 2y = 3) × 3
Or, 3x + 6y = 9 and 3x + y = 4
or, 5y = 5
or, y = 1 Putting the value of y = 1 in (i) then; x + 2y = 3
or, x + 2 × 1 = 3
Or, x = 1
So, the passing point is (1, 1).
Centre (h, k) = (1, 2)
We have, radius r2= (2 – 1)2 + (1 – 1)2 = 1 unit
Now, equation of circle is (x – h)2 + (y – k)2 = r2
or, (x – 1)2 + (y – 2)2 = 1
or, x2 – 2x + 1 + y2 – 4y + 4 = 1
or, x2 + y2 – 2x – 4y + 4 = 0
Thus, the equation of the circle is x2 + y2 – 2x – 4y + 4 = 0.

Find the equation of the circle having the centre (4, 6) and passing
through the midpoint of the line joining the points (– 1, 3) and (3, 1).
Solution:
Here, centre of circle (h, k) = (4, 6) and passing point is the midpoint of (–1 , 3) and
(3, 1).
So, passing point x =
x1 + x2
2
and y =
y1 + y2
2
Or, x =
−1+3
2
and y =
3+1
2
X= 1 and y= 2
Now, radius = distance between (1, 2) and (4, 6)
= (4 – 1)2 + (6 – 2)2 = 32 + 42 = 25 i.e. r= 5units
We know that, equation of circle is (x – h)2 + (y – k)2 = r2
or, (x – 4)2 + (y – 6)2 = 52
or, x2 – 8x + 16 + y2 – 12y + 36 = 25
or, x2 + y2 – 8x – 12y + 27 = 0
Thus, the required equation of circle is x2 + y2 – 8x – 12y + 27 = 0.

Find the equation of the circle having the centre as the point of
intersection of the lines x – y = 4 and 2x + 3y + 7 = 0 and passing
through the point (2, 4).
Solution:
Here, given equation of lines are, x – y = 4 ………. (i)
2x + 3y = – 7 …….. (ii)
Solving equation (i) × 2 and subtracting equation (ii)
then, 2x – 2y = 8
2x + 3y = – 7
– – + .
– 5y = 15
or, y = – 3
Putting y = – 3 in equation (i) then, x – y = 4 or, x + 3 = 4 or, x = 1
So, the centre of circle (h, k) = (1, – 3)
Radius = distance between (1, – 3) & (2, 4)
= (x2 – x1) 2 + (y2 – y1) 2 = (2 – 1)2 + (4 + 3)2 = (1)2 + 72
or r2 = 50
We know that, equation of circle with centre (h, k) and radius r is (x – h)2 +
(y – k)2 = r2
or, (x – 1)2 + (y + 3)2 = 50
or, x2 – 2x + 1 + y2 + 6y + 9 = 50
or, x2 + y2 – 2x + 6y = 40
Thus, the equation of circle is x2 + y2 – 2x + 6y – 40 = 0.

Find the equation of the circle passing through the points (2, 3) and (5, 4) and centre
on the line 2x + 3y – 7 = 0.
Solution:
Here, given equations of centre line is 2x + 3y = 7.
the circle passing through the points (2, 3) and (5, 4)
Let (h, k) be the centre of the circle then, 2h + 3k = 7 …. (i)
Distance between (2, 3) and (h, k) = Distance between (5, 4) and (h, k)
So, (h – 2)2 + (k – 3)2 = (h – 5)2 + (k – 4)2
or, h2 – 4h + 4 + k2 – 6k + 9 = h2 – 10h + 25 + k2 – 8k + 16
or, 6h + 2k = 28
Or, 3h + k = 14 …… (ii)
Solving equations (i) and (ii) × 3 then,
2h + 3k = 7
9h + 3k = 42
– – –
-7h = – 35
Or, h = 5

Cont…
Putting the value of h in (ii) then, 3 × 5 + k = 14 or, k = – 1
So, centre (h, k) = (5, – 1)
Radius = Distance between (2, 3) and (5, – 1)
i.e r2= (x2 – x1)2 + (y2 – y1)2
= (5 – 2)2 + (– 1 – 3)2 = 32 + 42 = 25
i.e. Radius = 5 units
Now, equation of circle is (x – h)2 + (y – k)2 = r2
or, (x – 5)2 + (y + 1)2 = 52
or, x2 – 10x + 25 + y2 + 2y + 1 = 25
or, x2 + y2 – 10x + 2y + 1 = 0
Thus, the equation of circle is; x2 + y2 – 10x + 2y + 1 = 0

On a wheel, there are three points (5, 7) , (-1, 7) and (5, – 1) located such that the
distance from a fixed point to these points is always equal. Find the coordinate of
the fixed point and then derive the equation representing the locus that contains
all three points.
Solution:
Here, let the given points be; A(– 1, 7), B(5, – 1) and C(5, 7).
Let P(h, k) be the point which is equidistant from the given points.
So, AP = BP = CP or, AP2 = BP2 = CP2
Taking AP2 = CP2 then, (h + 1)2 + (k – 7)2 = (h – 5)2 + (k – 7)2
or, (h + 1)2 = (h – 5)2
or, h2 + 2h + 1 = h2 – 10h + 25
or, 12h = 24
Or, h = 2
Again, taking BP2 = CP2 then, (h – 5)2 + (k + 1)2 = (h – 5)2 + (k – 7)2
or, (k + 1)2 = (k – 7)2
or, k2 + 2k + 1 = k2 – 14k + 49
or, 16k = 48
Or, k = 3

Cont….
So the point of equidistant (h, k) = (2, 3) i.e. the centre of circle = (2, 3)
For radius; the distance between (2, 3) and (5, 7) i.e. r2 = (5 – 2)2 + (7 – 3)2
= 32 + 42 =25
or, r = 5 units
Now, equation of the locus is; (x – h)2 + (y – k)2 = r2
or, (x – 2)2 + (y – 3)2 = 52
or, x2 – 4x + 4 + y2 – 6y + 9 = 25
Or, x2 + y2 – 4x – 6y + 13 = 25
Or, x2 + y2 – 4x – 6y = 12
Thus, the equation of the locus is x2 + y2 – 4x – 6y = 12.

Find the radius of the circle which passes through the points
A(– 4, – 2), B(2, 6) and C(2, – 2).
Solution:
Here, given points are A(– 4, – 2), B(2, 6), C(2, – 2)
We have equation of the circle is; (x – h)2 + (y – k)2 = r2 ..... (i)
Since equation (i) passes through the points; A(– 4, – 2), B(2, 6) and C(2, – 2).
So, (– 4 – h)2 + (– 2 – k) 2 = r2 ....... (ii)
(2 – h)2 + (6 – k)2 = r2 ....... (iii)
and (2 – h)2 + (– 2 – k)2 = r2 ...... (iv)
From equation (ii) and (iii)
Or, (– 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, 12h + 16k – 20 = 0
or,3h + 4k – 5 = 0 ....... (v)
Again from equation (iii) and (i)
(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, 16k – 32 = 0
or, 16k = 32
Or, k = 2

Putting the value of k in equation (v), we get,
Or, 3h + 4 × 2 – 5 = 0
or, 3h + 8 – 5 = 0
or, 3h + 3 = 0
or, 3h = – 3
or, h = – 1
Putting the values of h and k in equation (ii),
we get, [– 4 – (– 1)]2 + (– 2 – 2) 2 = r2
or, (– 4 + 1)2 + (– 4)2 =r2
or, (– 3)2 + 16 = r2
or, 9 + 16 = r2
or, 25 = r2
Or, r = 5
Thus, the required radius (r) = 5 units.

If the line x + y = 1 cuts a circle x2 + y2 =1 at two points,
find the distance between the two points.
Solution:
Here, given equation of line; x + y = 1 ........ (i)
Given equation of circle, x2 + y2 = 1
or, x2 + (1 – x)2 = 1 [From (i)]
or, x2 + 1 – 2x + x2 = 1
or, 2x2 – 2x = 0
or, 2x(x – 1) = 0
Hence, x = 0 or 1
When x = 0 then from (i) 0 + y = 1 i.e. y = 1
So, one point is (0, 1).
When x = 1 then from (i) 1 + y = 1 or, y = 1 – 1 i.e. y = 0
So, other point is (1, 0).
Now, the distance between two points; (0, 1) and (1, 0) is
d = (x2 – x1) 2 + (y2 – y1) 2 = (1 – 0)2 + (0 – 1)2 = 1 + 1 = 2 units.
Thus, distance between the two points is 2 units.

Find the equation of a circle concentric with the circle x2 + y2 – 6x +
y = 1 and passing through the point (4, – 2).
Solution:
Here, equation of circle is; x2 + y2 – 6x + y = 1
or, x2 + y2 – 6x + y -1=0………………………..(A)
Comparing it with general equation circle x2+y2+2gx+2fy+c=0
then we get, 2g= -6 i.e. g= -3 and 2f= 1 i.e. f= ½
Now we know that, center of a circle is (h, k)= (-g,-f) = (3,-1/2)
Equation of next circle centre (3,-1/2) and passing point (4, – 2) is :
Radius (r2)= (x2 – x1)2 + (y2 – y1)2 = (4 – 3)2 + (-2+1/2)2 = 1+9/4 = 13/4
Now, equation of circle is; (x – h)2 + (y – k)2 = r2
or, (x – 3)2 + (y+1/2)2 = 13/4
or, x2 – 6x + 9 + y2 + 2.y.1/2 + 1/4 = 13/4
or, x2 + y2 – 6x + y + 9 + 1/4 = 13/4
Or, x2 + y2 – 6x + y + 6 = 0
Thus, the required equation of circle is x2 + y2 – 6x + y + 6 = 0.

If the centres of two equal circles A and B are X and Y
respectively. If the coordinates of X is (2, 3) and the equation of
circle B is x2 + y2 – 2x + 6y + 1 = 0 then find the equation of the
circle A.
Solution:
Here, given equation of circle B is; x2 + y2 – 2x + 6y + 1 = 0
or, x2 – 2x + y2 + 6y + 1 = 0
or, x2 – 2 . x . 1 + 12 + y2 + 2 .y.3 + 32 = 12 + 32 – 1
or, (x – 1)2 + (y + 3)2 = 32
Comparing it with (x – h)2 + (y – k)2 = r2 ,
we get, radius (r) = 3 units which is also the radius of circle A.
So, in a circle A, radius (r) = 3 units and centre (h, k) = (2, 3)
We have the equation 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
x2 + y2 – 4x – 6y + 4 = 0
Thus, the required equation of circle A is; x2 + y2 – 4x – 6y + 4 = 0.

Find the equation of a circle with centre (3, 2) and passing through the centre
of the circle x2 + y2 – 2x + 4y – 4 = 0.
Solution:
Here, Equation of circle centre with (3, 2) is x2 + y2 – 2x + 4y – 4 = 0
Given equation of circle is; x2 + y2 – 2x + 4y – 4 = 0 ...... (i)
or, x2 – 2x + y2 + 4y – 4 = 0
Comparing it with general equation of circle x2+y2+2gx+2fy+c= 0
Then we get 2g= -2 i.e. g=-1 and 2f= 4 i.e. f= 2
We know centre of the circle (h, k) = (-g,-f) = (1, -2)
Which is the passing point of required circle.
Radius of required circle (r) = Distance (1, – 2) and (3, 2)
Or, r2= (3 – 1)2 + (2 + 2)2 = 22 + 42 = 20
Now, equation or circle is (x – h)2 + (y – k)2 = r2
or, (x – 3)2 + (y – 2)2 = 20
or, x2 – 6x + 9 + y2 – 4y + 4 = 20
Or, x2 + y2 – 6x – 4y – 7 = 0
Thus, the equation of circle is x2 + y2 – 6x – 4y – 7 = 0.

Find the equation of a circle with centre (– 1, 2) and passing through the centre of
the circle having equation x2 + y2 – 6x – 10y – 2 = 0.
Solution:
Here, given equation of circle is x2 + y2 – 6x – 10y – 2 = 0
Comparing it with x2 + y2 + 2gx + 2fy + c = 0,
we get g = – 3 and f = – 5
So, centre = (– g, – f) = (3, 5)
Now, radius of circle having centre at (– 1, 2) and passing point (3, 5) is,
Or, r2 = (3 + 1)2 + (5 – 2)2 = 16 + 9 = 5 units
Equation of circle having centre at (– 1, 2) and radius 5 units is: (x – h)2 + (y – k)2 = r2
or, (x + 1)2 + (y – 2)2 = 52
or, x2 + y2 + 2x – 4y + 5 = 25
or, x2 + y2 + 2x – 4y – 20 = 0
Thus, the equation of circle is, x2 + y2 + 2x – 4y – 20 = 0.

Determine the equation of the circle which passes through the points A(3, 0),
B(0, 3) & C(– 2, – 2).
Solution:
Here, let, C(h, k) be the centre and r be the radius of a circle so that the equation of the
circle is, (x – h)2 + (y – k)2 = r2 .................(i)
Since, the circle passes through A(3, 0), B(0, 3) and C(– 2, – 2)
⸫(3 – h)2 + (0 – k)2 = r2 ............(ii)
(0 – h)2 + (3 – k)2 = r2 ............(iii)
(–2 – h) 2 + (–2 – k)2 = r2 ........(iv)
From (ii) and (iii) we get,
Or, 9 – 6h + h2 + k2 = h2 + 9 – 6k + k2
or, – 6h + 6k = 0
or, k = h ..................(v)

From (ii) and (iv),
Or, 9 – 6h + h2 + k2 = 4 + 4h + h2 + 4 + 4k + k2
or, –10h – 4k + 1 = 0
or, 10h + 4k – 1 = 0 ...........(vi)
From (v) and (vi),
Or, 10h + 4h – 1 = 0
or, 14h = 1
or, h = 1/14 and k = 1/14
Now, r2 = (3 – 1/14)2 + (1/14)2
= (41/14)2 + (1/14)2
= (1681 + 1)/196
= 1682/196
= 841/98
Thus, (x – 1/14)2 + (y – 1/14)2 = 841/98 is required equation of
the circle.

Find the equation of circle passing through the points (3, 2) and (5, 4) and
having its centre on the line 3x – 2y = 1.
Solution:
Here, Let the coordinates of centre be (h, k).
Since the line 3x – 2y = 1 passes through (h, k) so, 3h – 2k = 1 .................. (i)
Since the circle passes through A(5, 4) and B(3, 2); PA = PB or, PA2 = PB2
or, (h – 5)2 + (k – 4)2 = (h – 3)2 + (k – 2)2
or, h2 – 10h + 25 + k2 – 8k + 16 = h2 – 6h + 9 + k2 – 4k + 4
or, – 10h + 6h – 8k + 4k + 41 – 13 = 0
or, – 4h – 4k + 28 = 0
or, – 4(h + k – 7) = 0
or, h = 7 – k .................(ii)
Putting the value of h in (i), 3(7 – k) – 2k = 1
or, 21 – 3k – 2k = 1
or, 5k = 21 – 1
⸫k = 4 Putting the value of k in (ii), h = 7 – 4 = 3
⸫Coordinates of centre of a circle is, (h, k) = (3, 4)
Now r2= (h – 3)2 + (k – 2)2 = (3 – 3)2 + (4 – 2)2 =4
Finally equation of circle r2= (x-h)2+(y-k)2
Or, 22= (x-3)2+(y-4)2
Thus, x2 + y2 – 6x – 8y + 21 = 0 is the required equation of circle.

Very Short Questions
1. Define the following terms. (a) Circle (b) Concentric circles
2. What is the equation of circle whose centre is (0, 0) and radius 'r'?
3. State the equation of circle whose centre is (a, b) and radius 'r'.
4. What is the equation of circle which passes through (0, 0) and has its centre at
(h, k)?
5. A circle has its centre at (0, 0) and radius 3 units. Write its equation.
6. A circle has centre at (– 4, – 5) and touches x-axis. What is its radius?
7. A circle has centre at (– 4, 2) and touches y-axis. Find is its radius.
8. A circle has radius 4 units and lies on fourth quadrant touching both axes.
Write the coordinates of centre.
9. State the diameter form of equation of circle.
10. What is the general equation of a circle?
11. Write the formulae to calculate centre and radius of a circle when its general
equation is given.
12. Write the equation of circle whose diameter has end points (a, b) and (p, q).
13. A circle has centre at (3, 4) and radius 5 units. Find its equation.
14. Find the equation of the circle whose
(a) Centre is (– 2, 5) and radius 17 units. (b) Ends of diameter are (3, 6) and (–1, 0).

Short Questions:
1. Find centre and radius of the following circles whose equation are;
(a) x2 + y2 = 25
(b) (x – 2)2 + (y – 3)2 = 36
(c) (x + 4)2 + (y + 5)2 = 14
2. If the line joining the points (1, 2) and (3, 6) is the diameter of a circle, find the equation of the circle.
3. If the line joining the points (2, 3) and (– 2, – 3) is the diameter of a circle, find the equation of the
circle.
4. If one end of the diameter of a circle x2 + y2 – 4x – 6y + 12 = 0 is (1, 3), find the co-ordinates of the
other end.
5. Find the equation of circle whose radius is 58 units and the centre lies on the x-axis and passing
through the point (5, 7).
6. Find the equation of circle whose radius is 13 units and the centre lies on the y-axis and passing
through the point (– 3, 9).
7. Find the equation of a circle whose centre is (– 3, – 4) and touches the X-axis.
8. Find the equation of a circle whose centre is (6, – 3) and touches the Y-axis.
9. Find the equation of the circle having centre at (4, 5) and touches the X-axis.
10. Find the equation of the circle having centre at (– 5, – 4) and touches the Y-axis.
11. Find the equation of circle whose radius is 4 units and touches both the axes lies on third quadrant.

Long Questions:
1. Find centre and radius of the following circles whose equation are;
a) x2 + y2 – 18x + 71 = 0
b) x2 + y2 – 2y = 24
c) 2x – 6y – x2 – y2 = 1
d) 2x2 + 2y2 – 2x + 6y = 45
e) 5x2 + 5y2 – 50x + 20y – 65 = 0
f) 2x2 + 2y2 + 6x – 2y = 13
2. Find the equation of the circle passing through the following points:
a) (0, 0), (2, 0) and (0, 4) b) (– 2, 2) (2, 4) and (4, 0) c)(1, 1), (4, 4) and (5, 1)
3. Find the equation of a circle which passes through the points (5, 5) and (3, 7) and
centre lies on the line x – 4y = 1.
4. Find the equation of a circle which passes through the points (4, 1) and (6, 5) and its
centre lies on the line 4x + y = 16.
5. The circle A with centre X passes through the centre Y of the circle B. If the equation
of circle B is x2 + y2 – 4x + 6y – 12 = 0 and the co-ordinates of X is (– 4, 5), then find the
equation of the circle A.
6. Find the equation of the circle whose centre is (4, 5) and it passes through the centre
of circle x2 + y2 + 4x + 6y – 12 = 0.
7. Find the equation of a circle concentric with the circle x2 + y2 – 6x + y = 1 and passing
through the point (4, – 2).

8. A circle has radius 5 units and centre is the point of intersection of the lines 2x – y = 5 and x – 3y +
5 = 0. Find the equation of circle.
9.Find the equation of a circle whose centre is the point of intersection of the line x – y = 9 and x + y
= 19 and which passes through the point (5, 8).
10. Find the equation of the circle which touches the x-axis at a point (3, 0) and passing through the
point (1, 2).
11. Find the equation of the circle with radius 5 units and which touches the positive x-axis and y-
axis.
12. Find the equation of the circle which touches both the positive axes and its centre lies on the line
6x + y = 14.
13. Find the equation of the circle which passes through the points (3, 2) and (5, 4) and its centre lies
on the line 3x – 2y – 1= 0.
14. Show that the two circles x2 + y2 = 36 and x2 + y2 – 12x – 16y + 84 = 0 touch externally.
15. Show that two circles touch internally. x2 + y2 = 81 and x2 + y2 – 6x – 8y + 9 = 0 .
16. The equations of two diameters of a circle passing through the point (3, 4) are x + y = 14 and 2x –
y = 4. Find the equation of the circle.
17. Find the equation of a circle whose centre is at the point of intersection of 2x + y = 4 and 2y – x =
3 and passing the point (4, 6).

Very long Questions:
1. On a wheel, there are three points (5, 7) , (− 1, 7) and (5, – 1) located such that the
distance from a fixed point to these points is always equal. Find the coordinate of
the fixed point and then derive the equation representing the locus that contains
all three points.
2. Three students are sitting in a playground to make a circular path. They are sitting
on the circumference, the co-ordinates of their position on the circle are (1, 2), (3,
4) and (3, 2). Find the co-ordinates of the point equidistant from their position.
Also, find the equation of the locus.
3. The line x – y = 0 cuts the circle x2 + y2 + 2x = 0 at two points A and B. Find the co-
ordinates of the points A and B. Also, find the equation of a circle whose diameter
is AB.
4. The line x – y + 2 = 0 cuts the circle x2 + y2 + 2x = 0 at two points A and B. Find the
co-ordinates of the points A and B. Also, find the equation of a circle whose
diameter is AB.
5. Find the equation of the circle which passes through the point (5, 8) and touches
the straight lines x = 1, x = 9 and X-axis.
6. Find the equation of the circle which touches the x –axis at (4, 0) and cuts off an
intercepts of 6 units from the y – axis positively.
7. Prove that the points A(2, –4), B(3, –1), C(3, –3) and D(0, 0) are concyclic.

Circle

More Related Content

What's hot

Similar to Circle

Recently uploaded

Circle