CIRCLES
Definition: A circle is the locus of a point which moves in a plane so that it is always at constant distance from a fixed point in a plane. The fixed point is called the centre of the circle
and
the constant distance is called the radius
of the circle.
(x,y)
FIG 1
Figure 1 : shows a circle of radius with centre at point O(a, b). The point (x, y) is on this circle if and if only if its distance from the centre c (a,b) is r, this means that a circle in the plane consist of all point (x,y), that are a given positive distance r from a fixed point (a, b).
One can express the relationship between centre (a,b), radius r can point (x,y) on the circle by using distance formula which will give.
(x–a)2 + (y – b) = r2 – – – (1)
The standard form of the equation of a circle
(x – a)2 + (y – b) = r2
Where the point (a,b) is the centre and r is the radius of the circle.
Example
Find the equation of the circle with centre (–2,5) and radius 4 unit
Solution
From (1) the equation
[x– (–2)]2 + (y – 5)2 = 42
(x+2)2 + (y – 5)2 = 16
x2 + 4x + 4 + y2 – 10y + 25 = 16
x2 + y2 + 4x – 10y + 13 = 0
CENTRE AT THE ORIGIN
If the centre of the circle is the origin (0,0) by the equation (x–a)2 + (y – b) = r2
(x – 0)2 + (y – 0) = r2
x2 + y2 = r2 – – – (2)
Example
Find in the simplest form the equation of the circle with centre 0 and radius
Solution
From (2), the equation of the circle
Is x2 + y2 = 5
Example
The point (–1,1) lies on a circle whose centre is at (3,–2) find the equation of the circle
Solution
The radius r of the circle is the distance between (–1, 1) and (3, –2)
r2 = (x – a)2 + (y – b)2
r2 = (–1–3)2 + [(1 – (– 2)]2
r2 = 25
r = 5 units
Thus, the centre of the circle is (a,b) = (3,2) and the radius is 5 units
Hence the equation of the circle
(x – 3)2 + [y – (–2)]2 = 52 (standard form)
x2 + 6x + 9 + y2 + 4y + 4 = 25
x2 + y2 – 6x + 4y – 12 = 0 (general form)
Example
The ends of a diameter of a circle are (5,6) and (7,8). Find the coordinates of the centre and length of the radius. Hence find the equation of the circle.
Solution
A(5, 6) (7,8)
(a) The centre of the circle is the midpoint of AB
Therefore the coordinates of centre is
(b) The radius of the circle is the distance between either (5,6) and (6,7) or
(6, 7) and (7, 8)
(c) The equation of the circle is
(x – 6)2 + (y – 7)2 =
x2 – 12x + 36 + y2 – 14y + 49 = 2
x2 + y2 – 12x – 14y – 83 = 0
Example:
Find the equation of the circle passing through the points (4,1) and (6,5) and having its centre on the line 4x + y =16
Solution
The slope of the line joining the points P(4,1) and Q (6,5) is
The slope of the of PQ is
The coordinate of the midpoint of PQ are i.e. (5,3)
The equation of the is
This equation (x + 2y = 11) intersect the line
4x + y =16 at the centre of the circle (a,b).
Solving x + 2y = 11 and 4x + y =16 simultaneously gives x = 3, y = 4
The centre of the circle is (a, b) is the point (3,4)
The radius of the circle is
The equation of the circle is
Example
Find the equation of the circle whose centre is the point (2,3) and the line 5x +2y – 7 = 0 is a tangent to the circle.
Solution
The perpendicular distance will give the radius of the circle
The equation of the circle is
EQUATION OF
The equation is represented as
Example
Obtain the equation of a circle which has the position A (1, 3) and B (4, 5) as the ends of its diameter.
SOLUTION
GENERAL FORM OF THE EQUATION OF A CIRCLE
The standard form of equation of a circle with centre C (a, b) and radius r is
(x – a)2 + (y – b)2 = r2 which gives
x2 + y2 – 2ax – 2by + a2 + b2 – r2 = 0 {Expand terms}
The above equation can be written as
x2 + y2 + 2gx + 2fy + c
Where a = –g, b = –f and c = a2 +b2 – r2
The equation x2+y2+2gx+2fy + c = 0 is the General form of the equation of a circle. Any of g, f, ,c, can be zero.
Note:– It can be seen that the general equation of a circle has the following characteristics
1. It is a general degree equation in x and y
2. The coefficients of x2 and y2 are equal
3. There is no term in xy
CENTRE AND RADIUDS OF A CIRCLE
The general form of the equation of a circle is of less usefulness then the standard form for instance; it is not immediately apparent form the general equation of the circle
x2 + y2 + 4x + 6y + 4 = 0; the radius and the coordinates of the center of the circle
[r = 3 unit, (a,b) = (–2, – 3)]
Example
Find the center and radius of a circle whose equation is 16x2 + 16y2 + 16x +40y – 7 = 0
Solution
To write the given general form in standard form, one must complete the square for x – term and y – term
16x2+ 16y2 + 16x + 40y – 7 = 0
x2 + y2 + x – = 0 {Divide through by 16 so as have x2 and y2}
(x2 + x + ) + (y2 –5y/2 + ) =
(half)2 (half)2
(half)2 (half)2
(x+3)2 + (y+2)2 = 9
The coordinate of the center is (–3, –2) and the radius is 3 units.
THE EQUATION OF A CIRCLE THROUGH THREE NON–COLLINEAR POINTS
To find the equation of a circle through three points
(a) Substitute the coordinate of each point in term in the general equation of the circle
(b) Solve the three simultaneous equations to find the value of g, f and c
(c) Substitute g , f and c in the general equation
Let the equation be x2 + y2 + 2gx + 2fy + c = 0
Example
Find the equation of the circle passing through points (–2, 5), (1, 4) and (5, –3)
Solution
Let the equation of the circle
x2 + y2 + 2gx + 2gy + c = 0
Since the given points lies on the circle
We have
at (–2, 5) (–2)2 + (5)2 +2(–2)g + 2(5)f + c = 0
29 – 4g + 10f + c = 0 – – – – – –– – – – – (i)
at (1, 4) (1)2 + (4)2 + 2(1)g+ 2(4)f + c = 0
17 + 2g + 8f + c = 0 – – – – – – – – – – – –(ii)
at (5, –3) (5)2 + (–3)2 + 2(5)g + 2(–3)f + c = 0
34 + 10g – 6f + c = 0 – – – – – – – – – – (iii)
Bringing out the three equations we have
29 – 4g + 10f + c = 0 – – – – – – –(1)
17 + 2g + 8f + c = 0 – – – – – – – (2)
34 + 10g – 6f + c = 0 – – – – – – ––(3)
Subtracting (1) from (2)
– 12 + 6g – 2f = 0 – – – – – – – – – – – – – – – – – –(4)
Subtracting (1) from (3)
5 + 14g – 16f = 0 – – – – – – – – – – – – – – – – – –(5)
Multiplying (4) by 8
– 96 + 48g – 16f = 0 – – – – – – – – – – – – – – – –– (6)
Subtracting (6) from (5)
101 – 34g = 0
Substitute for g in (4)
Example
Find the equation of the circle circumscribing the triangle formed by the lines
4y – 5x = 3, 7y – 4x = 10 and x + 3y = 26
Solution A(1,2)
4y–5x=3
7y–4x=10
B (5,7)
C (8,6)
x–3y=26
4y – 5x = 3 – – – – – – – – – (1)
7y – 4x = 10 – – – – – – – – –(2)
x + 3y = 26 – – – – – – – – – –(3)
From our knowledge of intersection of lines we can obtain the coordinates of each points
The coordinate of point A
4y – 5x = 3
7y – 4x = 10
The solving this simultaneously gives x = 1, y = 2
Coordinates of A = (1,2)
Coordinates of point B
4y – 5x = 3
x + 3y = 26
solving this simultaneously gives x = 5, y = 7
Coordinate of B = (5,7)
The coordinate of point C
7y – 4x = 10
x + 3y = 26
Solving this simultaneously gives x = 8, y = 6
The coordinate of C = (8,6)
Using the general equation of circle
x2+y2+2gx+2gy + c = 0
at A(1,2) 5+2g+4f+c = 0 – – – – – – –(A)
at B(5,7) 74 + 10g + 14f + c = 0 – – –(B)
at C(8,6) 100 + 16g + 12f + c = 0 – ––(C)
Subtract (A) from (B)
69 + 8g + 10f = 0 – – – – – – – (D)
Subtracting (A) from (C)
95 + 14g + 8f = 0 – – – – – – – – – – – – – – – – – – – – – – – – – – – (E)
Multiply (D) by 4 = 0
276 + 32g + 40f = 0 – – – – – – – – – – – – – – – – – – – – – – – – – – –(F)
Multiply (E) by (5)
475 + 70g + 40f = 0 – – – – – – – – – – – – – – – – – – – – – – – – – – (G)
Subtract (F) from (G)
19x2 + 19y2 – 199x – 103y + 310 = 0
EQUATION OF THE TANGENT AT THE POINT (X1Y1)
P(x1y1) (x, y)
Let (x1y1) and C(–g–f) or (a,b) be the center of the given circle. The tangent at P will be perpendicular CP. The slope of CP is and hence, the slope of the tangent at P is
(y – y1) (y1+ f) = – (x1+ g) (x – x1) – – – – – – – – – – – – – – – – – – – (1)
yy1 – y12 + fy – fy1 = –xx1 – gx + x12 + gx1
xx1 + yy1 + gx + fy = x21 + y12 + gx1 + fy1 – – – – – – – – – – – – –(2)
Adding gx1 + fy1 + C to each sides, the equation becomes
xx1 + yy1 + gx + fy + gx1 + fy1 + C = x12 + y12 + gx1 + fy + gx1 + fy + C
xx1 + yy1 + g(x+x1) + f(y+y1) + C = x12 + y12 + 2gx1 + 2fy1 + C – – –– – – – – – –(3)
the equation for (x1y1) which lies on the circle is x12 + y12 + 2gx1 + 2fy1 + C = 0
C = – (x12 + y12 + 2gx1 + 2fy1) – – – – – – – – – – – – – – – – – – – – –(4)
Substitute (4) in RHS of 3 we have xx1 + yy1 + g(x + x1) + f(y+y1) + C = 0
Hence the equation of the tangent to a circle
x12 + y12 + 2gx1 + 2fy1 + C = 0 at point (x1 y1) on the circle is
xx1 + yy1 + g(x+x1) + f(y+y1) + C = 0
Note: This equation can be obtained from the general equation of a circle by replacing
x2 by xx1, y2 by yy1; 2x by (x+x1) and 2y by (y+y1).
Example
Show that the point (3,1) lies on the circle x2 + y2 + 4x – 10y – 12 = 0
and hence, determine the equation of the tangent at (3,1) to the circle x2 + y2 + 4x – 10y – 12
Solution
Substituting the coordinate (3,1) into the equation
LHS = 32 + 12 + 4(3) – 10(1) – 12
= 9 + 1 + 12 – 10 – 12
= 22 – 22
= 0
= RHS
Hence the point (3,1) lies on the circle x2 + y2 + 4x – 10y – 12 = 0
Equation of the tangent at (x, y) is
xx1 + yy1 + g(x+x1) + f(y+y1) + C = 0
g = 2, f = –5
3x + y + 2(x+3) –5 (y+1) – 12 = 0
3x + y + 2x + 6 – 5y – 5 – 12 = 0
5x + 5y – 6 = 0
LENGTH OF THE TANGENT
Consider Length of the tangent from the point
A(x1y1) drawn to the circle x12 + y12 + 2gx1 + 2fy1 + C = 0
B A(x1y1)
C = (–g –f) is the centre of the circle
A = (x1y1) while line BA is the tangent A
Using Pythagoras theorem, the distance between C and A can be calculated
BA2 = AC2 – BC2
But AC2 = (x1+g)2 + (y1+f)2
BA2 = [(x1+g)2 + (y1+f)2] – [g2 + f2 –C]
BA2 = x21 + y12 + 2gx1 + 2fy1 + C
(which gives the length of the tangents)
Example
Find the length of the tangent drawn to the circle x2 + y2 + 5x + 4y – 20 = 0 from a point (2,3)
Solution
Equation of the circle is x2 + y2 + 5x + 4y – 20 = 0
The length of tangent from (2,3) to the circle is
Example
Find the length of the tangent drawn form a point (5,4) to the circle
4x2 + 4y2 + 8x + 5y + 16 = 0
Solution
Equation of the circle is 4x2 + 4y2 + 8x + 5y + 16 = 0
Divide through by 4
= x2 + y2 + 2x + y + 4 = 0
Example
Find the equation of the tangent at the point (2, –3) on the circle
4x2 + 4y2 – 6x + 9y – 13 = 0
Solution The equation of tangent to a circle xx1 + yy1 +g(x+x1)+f(y+y1)+ C= 0
4x2 + 4y2 – 6x + 9y – 13 = 0
Alternatively
We may not reduce the coefficient of x2 and y2 to unity as it was when the center and radius had to be found.
4x2 + 4y2 – 6x + 9y – 13 = 0
4x(2) + 4y(–3) –3(x+2) + (y – 3) – 13 = 0
8y + 12y – 3x – 6 + y – – 13 = 0
16x – 24y – 6x – 12 + 9y – 27 – 26 = 0
10x – 15y – 65 = 0
Example
The length of the tangent from the point (3, 2) to the x2 + y2 – 2x + 3y + C = 0 is 18 units. Find the value of C.
Solution
Length of tangent = x2 + y2 – 2x + 3y + C at point (3, 2)
(Squaring both sides) 324 = 9 + 4 – 6 + 6 + C
C = 311
Example
The length of the tangent from the point (2 ,–1) to the circle 3x2 + 3y2 + 4x + 2y + k = 0 is 3 units. Find the value of K
Equation of circle is 3x2 + 3y2 + 4x + 2y + k = 0
SOLUTION
Dividing through by 3 to reduce to the coefficient of x2 and y2 to unity
x2 + y2 +
Length of the tangent
THE POINT OF INTERSECTION OF THE STRAIGHT LINE y = mx + c AND THE CIRCLE x2 + y2 = r2
The equation of the circle with center 0
x2 + y2 = r2 – – – – – – – – –– – – – – – (1)
the line y = mx + C – – – – – – – – – (2)
Since the points of intersection lies on both the line and the circle, they will be given by the solution of equation (1) and (2)
Solving the two equation
Substituting (2) into (1)
x2 + (mx + c)2 = r2
x2 + m2x2 + 2mcx + C2 = r2 – – – – – – – – – – – – – (3)
Equation (3) is a quadratic in x and will give two values for x and from (1), the two corresponding values of y can be obtained
We have the following cases
CASE 1
Equation (3) has real and distinct
(2mc)2 – 4(1+m2) (c2– r2) > 0
r2(1+m2) – c2 > 0
r2 (1+m2) > c2
Wherever r2 (1 + m2) > c2 equation (3) has real and distinct roots, meaning the line cuts the circle in two real and distinct points.
CASE II
Equation (3) has equal roots
The condition for this is
r2(1 + m2) = c2
In this case the two points of intersection are coincident, Hence, the line y = mx + C
are tangents to be circle x2 + y2 = r2 for all values or m,
is defined as the slope equation of the tangent.
CASE III
Equation (3) has imaginary roots. The condition for this will be r2(1 + m2) < C2
In this case the points of intersection are said to imaginary
Example
Find the equation of the tangents to the circle x2 + y2 = 25
1 Which are inclined at angle of 600 to x – axis
2. Which are parallel to the line 4x – y + 8 = 0
3. Which are perpendicular to the line 5y = 5 – 4x
Solution
(1) Tangent to the circle x2 + y2 = 25 which is incline at angle of 600 to x–axis
r = 5, slope = m = tan 60 =
Hence the required tangent y = mx + r
(2) Tangent to the circle x2 + y2 = 25 which is parallel to the line 4x – y + 8 = 0
Slope = m = 4 {m0 = m condition for parallelism}
y = 4x + 5
y = 4x + 5
(3) Tangent to the circle x2 + y2 = 25 which is perpendicular to the line 5y = 5 – 4x
m0 = – 4
Slope of the tangent m = ¼ {m1m2 = – 1 (condition for perpendicularity)}
Example
Find the tangent to the circle x2 + y2 = 10 from the point (4 – 2)
Solution
Any line through the point (4, –2) with slope m is y + 2 = m (x – 4)
y + 2 –(mx – 4m) = 0
The line y + 2 –(mx – 4m) = 0 will be will tangent to circle x2 + y2 = 10, The radius of the circle is equal to the length of the perpendicular from the center (0, 0) on the line (a)
i.e. the distance of (x1 + x2) from the line y + 2 –(mx – 4m) = 0 is +
(4m + 2)2 = 10 (m2 + 1)
16m2 + 16m + 4 = 10m + 10
6m2 + 16m – 6 = 0
(6m – 2) (m + 3) = 0
m = or m = –3
Hence the two tangent are
y + 2 = (x – 4) or x – 3y = 10
y + 2 = – (x – 4) or 3x + y = 10
Example
Show that x2 + y2 – 11x – 7y + 30 = 0 and x2 + y2 –7 x + 3y – 110 = 0 touch one another at point (8,6). Find the equations of the tangents to each other of circles at the point (8, 6) and hence deduce that the circles touch each other
Solution
Since 82 + 62 – 11(8) – 7 (6) + 30 = 0
And also 82 + 62 – (8) – 7(6) + 3(6) – 110 = 0
This show that the point (8, 6) lies on both the circles.
Equation of tangent for the 1st circle x.8 + y.6 – = x + y – 14 = 0
Equation of tangent for 2nd circle is
thus we find that two circles passes through point (8, 6) and also have the same tangent at point (8, 6) implying that the circles touches each other at (8, 6)
Example
Find the points of intersection of the line x + y =3 and the circle x2 + y2 + x –5y + 4 =0
Solution
The points of intersection is gotten by solving the two equations.
x = 3 – y ---------------(i)
(3 – y)2 + y2 + (3 – y) – 5y + 4 = 0 {substitute (3 – y) =x into equation of the circle
y2 – 6y + 9 + y2 + 3 – y – 5y + 4 = 0
2y2 – 12y + 16 = 0
y2 – 6y + 8 = 0
(y – 4) (y – 2) = 0
y = 4 or y = 2
When y = 4, x = –1
When y = 2, x = 1
Thus, the two points of intersection are
(–1,4) and (1,2)
Example
Show that the line 3x – 4y – 10 = 0 is a common tangent of two circles x2 + y2 = 4 and
x2 + y2 – 22x – 24y + 240 = 0
Solution
The points of intersection is gotten for both circles by the solving the equation
3x – 4y – 10 = 0
In the circle x2 + y2 = 4
In the circle x2 + y2 – 22x – 24y + 240 = 0
Using the equation of line
Therefore the line 3x – 4y – 10 is a common tangent to two circles.
FAMILIES OF CIRCLES
INTERSECTING CIRCLES
The equation of any circle the intersection of circles
That is equation of a straight line, which passes through the point of intersection of the two circles (i.e. common chord). And is known as the radical axis of the circles
Example
Find the equation of the circle which passes through the point(1,2)and the point of intersection of circles
and
Solution
A circle through the intersection of the given circles has equation of the form
Since the point (1, 2) must lie on
this circle
So the required equation is
Example
Find the equation of the family of circles through the intersection of the two circles and having its centre on the x – axis.
Solution
A circle through the intersection of the given circles has equation of the form
or
Since emerging circle has one has its centre on the x – axis, that is f =0
The required circle is
Family of concentric Circle
A family of circles with same centre
and different radii is called family of concentric circle
If centre is given, then two constant g and f in the equation
are determine because centre of circle is (–g, –f)
An infinite number of circles can belong to the family of concentric circles.
Example
Find the equation of the circle concentric to the circle and which passes through the origin
Solution
Let the family of circles concentric to be
From the given equation (-g ,-f ) = (3,0)
Since the point (0, 0) must lie on this circle
Radius of new circle =
Equation of the new concentric circle will be
Example
Find the equation of the circle which is concentric to the circle and touches the line x + y + 3 = 0
Let the family of circles concentric to be
From the given equation (-g ,-f ) = (3,0)
Radius of new circle =
Equation of the new circle
Orthogonal Circles
Two curves are said to intersect orthogonal, when thy intersect at right angles
To determine the condition for two circles to intersect orthogonally
Let the equation of the intersecting circles be
The coordinates of the centre are (– g1, – f1 )
and (– g2, – f2 )
The radii of the circles are and
The condition that the circle cut orthogonally at C (i.e ÐACB is a right angle ) is
AB2 =AC2 + CB2 , that is
The condition is
Example
Prove that following circles cut orthogonally
Solution
Coordinate of the centre C1 are (0,0), c1 = –25
Coordinate of centre C2 are (0,13), c2 = 25
These circles will cut orthogonally if
2(0)(0) + 2(0)(0) = 25 – 25 which is true 0 = 0
Hence the given circle cut orthogonally
ANGLE OF INTERSECTION OF TWO CIRCES
The angle of intersection of two curve is defined to be the angle bewteen the tangent to the curves at their points of intersection
Let r1 and r2 be radii of the two circles , d the distance between their centres and q the angle between them
The above formula helps us to determine the angle between two circles
Example
A circle touches the x –axis and the line 4x –3y + 4 = 0. Its centre is in the first quadrant and lies on the line x –y –1 = 0. Prove that the equation of the circle is
Solution
Let the point A where the circle touches the x –axis be (x1, 0).
Therefore equation of the perpendicular to the tangent lying on the x-axis is x = x1
Since this perpendicular passes through the centre of the circle x = x1 gives the x – coordinate of the centre
x – y –1 = 0 is a line passing through the centre
y = x – 1 = x1 – 1.
Coordinate of the centre (x1, x1 –1)
Gradient of
the line –3y + 4x +4 = 0 is 4/3. Hence, the equation of line
Therefore B is obtained by solving and simultaneously. This gives
Click here to go back to the content page