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CIRCLE
Is the locus which sown on xy – plane so that it always constant distance i.e. radius from fixed pilot i.e. Centre.
− Where the phiral of radius is radii.
GENERAL EQUATION OF CIRCLE
− consider the figure below on X – Y plane
Square both sides
r² = 
Hence
r² =
The general formula for equation of the circle.
− By extending the above formula we get the general equation of circle.
= r²
² – 2ax + a² + y²- 2by + b² = r²² + y² – 2ax – 2by + a² + b² – r² =0² + y² + 2
y + a² + b² r² =0
-a = 
-b =
Therefore
+ y² + 2x + 2
Show as equation of circle
r
But ‾a = g, a= ‾g
‾b = f, b = ‾f
Also the coordinate of the center of the culve ( –g, –f)
1. To find the Coordinate of the Centre and radius of the circle make sure coeficient of X2 and y2 is equal to 1.
2. Centre of the culve is (-1/2 coefficient x) (-1/2 coefficient + y)
3. If the circle 
posses through the origin, then c =0
posses through the origin, then c =04. If the Centre lie on X- axis f = 0
5. If the Centre lie on Y – axis g = 0
6. For the center at which the coordinate of the center lie at origin be (90). Their general equation being in the form of 
DETERMINATION OF CENTRE AND RADIUS OF THE CIRCLE.
Mostly done by using the two common methods:
1. By using general equation of circle.
2. By using completing the square method.
I. By using general equation of circle
Here normally the general equation applied in order to get the value of G, F and C by comparing general equation of circle provided and equation of circle.
applied in order to get the value of G, F and C by comparing general equation of circle provided and equation of circle.Where,
− coordinate of Centre (-g, -f)
II. By completing the square method
− is another method at which performed under the same rule of completing the square method of a circle in order to express it in form of (x – a)2 + (y – b)2 = r2
Where,
i. Coordinate of the Centre taken under opposite sign of part of X and Y.
i.e. (x – 2)2 + (y + 3)2 = 9
C (a, b) = (2, -3)
ii. radius regarded as the square root of part of r2
r2 = 9
r2 
= 3.
= 3.EXAMPLE.
1. Determine the coordinate of the Centre and radius of the following
(a) x2 + y2 – 4x – 6y – 12 = 0
(b) 4x2 + 4y2 – 20x – 4y + 16 = 0
2. Write down the standard equation of the circle with Centre at the origin and whose radius is 5 units.
3. Find the equation of circle passing through (2, 1) and Centre at (-3, -4).
EQUATION OF THE CIRCLE WITH TWO POINTS AS THE END OF DIAMETER.
− Let AB b diameter of the circle with coordinate A (x1,y1) and B (x2, y2) which give general equation of (x – x1)(x – x2) + (y – y1)(y – y2) = 0.
Proof considers the figure below with two points as diameter.
Required equation of the circle let C be angle formed in the semi – circle.
Where < ABC = 90°
Since ACB = 90º
Then
Slope AB x Slope BC = –1
MAC x MBC = –1
Slope AB x Slope BC = –1
MAC x MBC = –1
But
MAC =
MAC =
MBC =
+
=0
+ 
=0EQUATION OF THE CIRCLE WITH THREE POINTS ON THE CIRCLE.
* Let consider the circle with three points on the circle such as A, B and C as shown below.
Generally the equation of the circle above calculated under the following methods.
I. By using general equation of the circle.
II. By using distance formula.
I. BY GENERAL EQUATION.
− Done by referring general equation of the circle x2 + y2 + 2gx + 2fy + c = 0
This form three equations in term of; f, g and c by substituting the point provided.
− Letters solves the equations three above in order to get the value of standard equation.
II. BY DISTANCE FORMULA
By using distance formula we formed two equations in term of A and B as C (a,b) under the reference of distance formula above from the Centre to the point on the circle.
e.g.
This form three equations in term of; f, g and c by substituting the point provided.
− Letters solves the equations three above in order to get the value of standard equation.
II. BY DISTANCE FORMULA
By using distance formula we formed two equations in term of A and B as C (a,b) under the reference of distance formula above from the Centre to the point on the circle.
e.g.
Where, AO = BC and BO = OC.
By solving formed equation we can get value of C (a, b) together with radius we can get the required equation.
EQUATION OF CIRCLE WITH TWO POINTS AND LINE PASSING AT THE CENTRE.
❖ Let two point be A (x1, y1) and B (x2 ,y2) and line L1 Px + Qy + c = 0 passing at the Centre.
Now we need to find the equation of circle
Important steps
i. Since the line passing at the Centre means that the points of Centre satisfy the line.
By substituting the point in the line px + qy + c = 0.
ii. Form second equation by using distance formula in terms of a’ and b’
I.e. AC = CB
iii. By solving equations as can get the Centre
EXAMPLES.
1) Find the equation of the circle on the line forming the points (-1, 2) and (-3, 5) as the diameter.
A circle is drown with points whose coordinate are A (OA) and B (P, Q) as the end of diameter, the can lie cuts X – axis two points where coordinate are (
) and (β, O), prove that + β = P and
A circle is drown with points whose coordinate are A (OA) and B (P, Q) as the end of diameter, the can lie cuts X – axis two points where coordinate are (
) and (β, O), prove that 
+ β = P and 1. Given
Points
B
B
Recall
=0x² + 3x + x + 3 + y² – 5y – 2y + 10 =0
x² + 4x + 3 +y² – 7y + 10 =0
x² + y² + 4x – 7y + 13 =0
x² + y² + 4x – 7y + 13 = 0From and (β,O)
and (β,O)Then x = β = X
β = X
(X – β) = 0X – Xβ – X + β = 0
– Xβ – X + β = 0x² – (∝+β) x + β =0
β =0x² – (∝+β)X + ∝β = 0
x – Px + a =0
– Px + a =0Then
–
+ 
= P
=a+ =aA= a
= aPoints A(‾1, ‾5) B(6, 2) and (0, 2)
By general equation
x2 + y2 + 2gx + 2fy + c =0
For A (‾1, 5)
26 – 2g – 10ƒ + c= 0
2g + 10 – c = 26 ——— (i)
+ + 2g + 2
36 + 4 + 12g + 4f + c = 0
12g + 4f + c = ‾40
For c
4 + 4f + c =0
4f + c =‾4 ————- (iii)
Hence
x2 + y2 + 2gx + 2fy + c =0
For A (‾1, 5)
26 – 2g – 10ƒ + c= 0
2g + 10 – c = 26 ——— (i)
+ 
+ 2
g + 2
36 + 4 + 12g + 4f + c = 0
12g + 4f + c = ‾40
For c
4 + 4f + c =0
4f + c =‾4 ————- (iii)
Hence
G=‾3, f = 2, c = ‾12
X² + y² + 2gx + 2fy + c =0
+ y² + 2 x + 2y – 12 = 0
+ y² + 2 
x + 2
y – 12 = 0x² + y² -6x + 4y– 12 =0
Equation the circle is x² + y² – 6x + 4y – 12 =0
Equation the circle is x² + y² – 6x + 4y – 12 =0EQUATION OF CIRCLE WITH COORDINATE OF CENTRE AND EQUATION OF TANGENT LINE.
Let consider Centre C (a, b) and tangent line px + qy -c = 0
Let consider Centre C (a, b) and tangent line px + qy -c = 0
Our intention is to find the equation of the circle.
IMPORTANT STEPS.
Determine the value of radius as shortest distance from Centre to the line.
By using value of radius and Centre we can get the equation.
Determine the value of radius as shortest distance from Centre to the line.
By using value of radius and Centre we can get the equation.
TO THE CIRCLE.
Equation of tangent to the circle at the given point regarded in order to form of equations such as.
Our intention is to find the equation of the tangent, let us find the slope.
By calculus method.
From
x² + y² = r²
But
of a curve at 
But
of a curve at 
Apply both sides
both sidesThen
2x + 2y
=0
2x + 2y
=02y= ‾2x
= ‾2x
= M=
Then
=
= 
a) By geometric method x² + y² = r²
Since
MCT ML = ‾1
But
MCT =
MCT =
ML = 
ML =
MT =
MT =
But equation of tangent
Y = M(x -x1) + y1
But M = 
= 
PROBLEMS.
1. Find the equation of circle with Centre (3,-4) and tangent to line 3x + 4y + 3 = 0.
2. Find the equation of the circle whose Centre is 9 units to the right of Y – axis whose radius is 2 units and which touch the line 2x + y – 10 = 0.
units and which touch the line 2x + y – 10 = 0.3. Find the equation of tangent to the circle x2 + y2 = 18 at pint (7, -3).
4. Find the equation of tangent for the circle through point (2,-4) given that x2 + y2 – 2x – 4y + 1 = 0.
5. Show that the point (7, -5) lies on the circle x2 + y2 – 6x + 4y – 12 = 0. Find the equation of tangent to the circle at the point.
5. Show that the point (7, -5) lies on the circle x2 + y2 – 6x + 4y – 12 = 0. Find the equation of tangent to the circle at the point.
Solution 1
C
From 3x + 4y + 3=0
From equation of circle
+
= r²
=
+ = 
Centre 9 units to the right y-axis radius = 2
(i) 2x + y – 10=0
2 x 5 = 8 + y
10 – 8 = y
y = 2
Then c(9,y)
But y =2
C(9,2)
From
= r²3. Given
x² + y² =58
Point
Point
Recall
xx1 + yy1 = r²
But (x1, y1) =
r² = 58
7x – 3y = 58
7x -3y =58
Solve equation 4
Given
Given
x² + y² – 2x – 4y + 1 =0
By using calculation approach
5. Give:solve equation 5.
Point (
7, -5)
7, -5)
x² + y² – 6x 4y – 12 =0
THE CONDITION OF CERTAIN LINE TO BE TANGENT TO THE CIRCLE.
In order y = Mx + c to be tangent to the circle x2 + y2 = r2, x2 + y2 + 2gx + 2fy + c = 0. always can be checked by the following method:
By using value of radius as perpendicular ions hence from Centre to the line y = Mx + c by solving y = Mx + C and x2 + y2 2gx + 2fy + c = 0. By substituting the value of Y into either of the equation of circle hence if b2 = 4ac, then the line is length of the circle.
In order y = Mx + c to be tangent to the circle x2 + y2 = r2, x2 + y2 + 2gx + 2fy + c = 0. always can be checked by the following method:
By using value of radius as perpendicular ions hence from Centre to the line y = Mx + c by solving y = Mx + C and x2 + y2 2gx + 2fy + c = 0. By substituting the value of Y into either of the equation of circle hence if b2 = 4ac, then the line is length of the circle.
Examples.
1) Find the value of K if 12x + 5y + K = 0. is length to the circle x2 + y2 – 6x -10y + 9 = 0
2) Show that 5x + 12y – 4 = 0. Touches the circle x2 + y2 – 6x + 4y + 12 = 0.
Solution
From
X² + y² – 6x – 15y + 9=0
c(-g,-f)=(-1/2(x),-1/2(y))
= 
Also r = 
r =
r = 5 units
r = from Centre to tangent
r =
at (x, y)
from Centre to tangentr =
at (x, y)
But r = 5
5 = 
5=
5 x 13 = 61 + k
65 – 61 = k
K = 4
The value of k = 14
The value of k = 14DISTANCE OF TANGENT FROM EXTERNAL POINT TO THE CIRCLE.
Consider the figure below.
By Pythagoras theorem
Consider the figure below.
By Pythagoras theorem
= 
+
= 
But
CP =
²Also
CT = radius
Then = g² + f² -c
= g² + f² -cFrom = ² –
= 
² –
² = 
+ 
–
= x² + 2 x g + g+ y² + 2fy + f² -g² – f² + c= x²+ 2 x g + y² + 2fy + c= x² + y² + 2xg + 2fy + c
=
EQUATION OF TANGENT TO THE CIRCLE FROM EXTERNAL POINT.
− Here the equation of tangent may be answered as Y = M (x1 – x1) + y1 where x, and y, are the external point which may be at the origin but not at contact of tangent to the circle.
− So in order to get equation we need to find the slope ‘M’ e.g. let consider equation of tangent from P(a, b) to the circle x2 + y2 = r2 or x2 + y2 + 2gx + 2fy + c = 0.
Y= M(x – x1) + y
Y= Mx – M1 + Y1
Mx– y – Mx1 + y1
From
r=d = 
r=
d = at
r=d = 
at
r=
d = 
at 
r = 
at
Hence the equation of the line can be obtained by using slope M from above formula together with point from.
GENERAL EQUATION OF TANGENT TO THE CURVE WITH THE GIVEN SLOPE.
Let Y = Mx + c be equation of tangent to the curve x2 + y2 = r2 or x2 + y2 + 2gx + c = 0. With given slope M to solve for c.
From
y= Mx + c
x² + y² = r²
Solve the above equation
x² + y² = r²
But y = mx + c
x2(mx+C)2=r2
y= Mx + c
x² + y² = r²
Solve the above equation
x² + y² = r²
But y = mx + c
x2(mx+C)2=r2
x² + m²x² + 2xmc + c²=r2
x²+m²x² + 2xmc + c² – r²=0
=0
a= m² + 1
b= 2mc
c= c2 – r2
Recall
b² =4ac
= 4
4m²c²= 4
4m²c²= 4m²c² – 4m²r² + 4c² 4r²
4m²c² = 4
m²c²= m²c² – m²r² + c² -r²
0= ‾m²r2 + c² – r²
c²= m²r² + r²
c² = r²
c=
c= 
Recall
y = Mx + c
y =Mx ± r
y = Mxr
y = Mx + c
y =Mx ± r
y = Mx
rPOSITION OF POINT WITH RESPECT TO CIRCLE
The done under the following
If then the point R lies inside the circle
If= r then the point P lies on the circle
then the point R lies inside the circle 
If
= r then the point P lies on the circle
If OR r then the point R we outside the circle
r then the point R we outside the circlePROBLEMS.
1. Find the length of tangent from (1, 2) to the circle, x2 + y2 – 4x – 2y + 4 = 0.
2. Find the length of tangent from the origin to the circle x2 + y2 – 10x + 2y + 13 = 0.
3. Find the equation of tangent from (-1, 7) to the circle x2 + y2 = 5.
4. Find the equation of tangent to the circle x2 + y2 = 16. If the slope of tangent is 2.
5. Examine whether the point (2, 3) lie at side/ inside the circle.
6. discuss the position of point (1, 2) and (6, 0) with respect to circle x2 + y2 – 4x – 2y – 11 = 0
Solution 1.
Given
(1, 2)
x² + y²- 4x – 2y +4 =0
c(-g, f) = (2, 1)
r=
r=
r=
r = 1units
Consider the figure below
By Pythagoras theorem
=2Also
= Radius
PT = 1 Unit of length
Length of tangent = 1 unitORTHOGONAL CIRCLES
This circle said to be orthogonal only if their radii are perpendicular to each other.
Simply their radii meet at right angle.
Means that circle I and II are orthogonal
The condition for orthogonal circles mostly explained by concept of Pythagoras theorem r12 + r22 =
Where
C1C2 = distance at Centre
r1 and r2 are radii of the circles.
Alternatively
From
r1
² + r2² =
But r1 = 
² + r2² =
But r1 = 
r1²= g1² + f1² – c ———- (i)
———- (i)Also
r2 =
= g2² + f2² – c2 ——– (ii)Let C1 , (-g1 , -f1) and C2 (-g2, -f2)
Recall
Recall
= 
= 
+ 
² —– (iii)From
r12 + r22 = 2²
2²
When and 2 are constant of the ci
rcle
and 
2 are constant of the circle
EXAMPLE
1. Find the value of K if circle x2 + y2 – 2y – 8 = 0 and x2 + y2 – 24x + Ky – 9 = 0 are orthogonal.
2. show that the following circle are orthogonal x2 + y2 – 6x -8y + 9 = 0
and x2 + y2 = 9.
3. Find the equation of circle which passes through the origin and cut orthogonally circle
x2 + y2 + 8y + 12 = 0 and x2 +y2 – 4x – 6y – 3 = 0.
4. find the equation of circle which cut the circle x2 + y2 – 2x – 4y + 2 = 0, x2 + y2 + 4x = 0.
5. find the equation of the circle through the point (2,0); (0, 2) and 2x2 + 2y2 + 5x–6y + 4 = 0
Solution 1
given
x² + y² – 2y – 8 =0
x² + y² – 24x + ky – 9 =0
Recall
2g1g2 + 2f1f2 =c1 + c2
Then
-g1 =0 = g1 =0
-f1 = 1 = f1 = ‾1
Also
-g2 = 12
g2 = ‾12
= 
f2 =
But C1 = -8, and C2 = ‾9
C1 + C2 = 2g1g2 + 2f1f2
‾8 – 9 = 2 + 2
+ 2
‾17 = ‾k
K = 17
The value of K = 17Solution 2
= 9From centres are
C1 and C2
and C2Constants
C1 = 9, C2= ‾9
C1= g1= ‾3, f1 = ‾4
= g1= ‾3, f1 = ‾4C2 = g2 = 0, f2 = 0
= g2 = 0, f2 = 0Then
C1 + C1= 2g1g2+ 2f1f2
9 -9 = 2 + 2
+ 2
0 = 0
Since LHS = RHS hence the circles are orthogonal Solution 3
Passing through the origin
C = 0
x² + y² -8y + 12 = 0 and
x² + y² – 4x – 6y – 3 =0
Let x² + y² + 2gx + 2fy + c =0
2g1 = 2g
g1 =g
2f1 = 2f
f1 =f
c1 = c
But c = 0
For 2nd equation
2g2 = 0
g2 =0
2f2 = ‾8
f2 = ‾4
c2 = 12
Recall
c1 + c2 = 2g1g2 + 2f1f2
2g + 2f = 12 +0
+ 2f
= 12 +0‾8f = 12
f=
Also let 2nd equation be
2g2 = ‾4
g2 = ‾2
2f2 = ‾6
f2 = ‾3
c2 = ‾3
Recall
2g1g2 + 2f1f2 = c1 + c2
But
g1=g ,
But
g1=g ,
f1 =f
2g + 2f = 0 +
+ 2f = 0 + ‾4g + 6f = ‾3
From
4g + 6f = 3
But f =
4g = 3 + 9
= 
g= 3
Value of g =3, f=
Recall
+ 2gx + 2fy + c =0
+ 6x – 3y =0The equation is + 6x – 3y =0
+ 6x – 3y =0Solution 4
Given
Points and
and 2x² + 2y² + 5x – 6y + 4 =0
Let
+ 2gx + 2fy + c = 0Point =
= 
4 + 0 + 4g + 0 + c =0
4 + 4g + c =0
4g + c = ‾4 ——– (i)
Also point (OR)
0 + 4 + 0 + 4f +C =0
4 + 4f + c =0
4f + c =‾4 ————-(ii)
Let
+ 2gx +2fy +c = 02g1 = 2g
g1=g
2f1 = f
f1 = f and c1 =c
The 2nd equation
2x² + 2y² + 5x – 6y + 4 =0
Recall
C1 + C2 = 2f1f2 + 2g1g2
C + 2= 2f + 2g
+ 2g
C + 2 = –3f +
– 3f – c = 2
g=
f=
c=
Then
+ 2gx + 2fy + c =0Substitute the value of g, f and c
x² + y² +x + 2y – =0
x + 2
y – 
=0
– 
–=07x² + 7y² – 8x – 8y -12 =0
The equation of the circle is 7x² + 7y² – 8x – 8y -12=0INTERSECTION OF TWO CIRCLES.
Intersection of two circles may be forced into three cases.
1. Common tangent.
2. common chord
3. Line of separation.
1) COMMON TANGENT.
Is a line at which two circle intersection at a single point.
2) COMMON CHORD.
Common chord is the line at which two circles intersection at two distance point.
3) LINE OF SEPARATION.
Line of separation is the line between two circles which has no point of intersection
How to find common tangent, chord or line of separation?.
This is done by either fut. strait the equation of one circle from another circle. i.e. C1 – C2 or C2 – C1
The different between two equations of circle represent common tangent, chord or line of separation.
Where,
The point of intersection of either common tangent or common chord obtained by solving them simultaneously under the following.
The point of intersection of either common tangent or common chord obtained by solving them simultaneously under the following.
1) If b2 = 4ac, the line is common tangent.
2) If b2 > 4ac, the line common chord.
3) If b2 < 4ac, the line is line of separation.
Note:
The other concept of intersection of circle explained as follow.
1) If the circle intersection externally.
CONCY CLIC CIRCLE.
* These are two circles which passes the same coordinate of Centre but varies is radius
Where
If four points are concyclic imply that by forming the equation of concyclic circle by using three points the fourth point should be satisfy the equation.
EXAMPLE.
1. Show that the part of the line, 3y = x + 5 is a chord of a circle, x2 + y2 – 6x – 2y – 15 = 0.
2. Find the equation of common chord and the intersection point of the circles x2 + y2 + 6x – 3y + 4 = 0, 2x2 + 2y2 – 3x – 9y + 2 = 0.
3. Find the length of the chord of the circle x2 +y2 – 2x -4y – 5 = 0, whose mid-point (2, 3).
4. Show that the circle x2 + y2 – 4x + 6y – 10 = 0, and x2 + y2 – 10x + 6y + 14 = 0and passing other.
5. Find the equation of the circle concentric with the circle x2 + y2 + 4x +6y + 11 = 0. And pass through.(5, 4)
6. Find the equation of circle which passes through the Centre, x2+ y2 + 8x + 10 y – 7 = 0, and is concentric with the circle 2x2 + 2y2 – 8x – 12y – 9 = 0.
7. Find the equation of the circle concentric with circle. 2x2 + 2y2 + 8x +10y – 39 = 0. And having its area equal to 16
Solution
Given
x² + y² + 4x + 6y + 11=0
Point
= 
R =
R =
R =
R=
(a, b) = (‾2, ‾3)
+ 
= r²+ = 98
