EQUATION OF CIRCLE
Definition:
A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle.
Equation of the circle with centre (h, k) and radius ‘r’ units.
CP = r
√(( x – h )2 + (y – k )2 = r (Using distance formula)
(x – h )2 + (y – k )2 = r2
Note:
The equation of the circle with centre (0, 0 ) and radius ‘r’ units is x2 + y2 = r2
Part – A
Example : Find the equation of the circle with centre (- 5, 7 ) and radius 5 units.
Soln: We know that the equation of circle is (x – h )2 + (y – k )2 = r2
Here h = – 5, k = 7 (given) and r = 5
. . (x + 5 )2 + (y + 7 )2 =
x2 + 10x + 25 + y2 + 14y + 49 = 25
x2 + y2 + 10x – 14y + 25 +49-25 = 0
x2 + y2 + 10x – 14y + 49 = 0
Therefore the equation of circle is x2 + y2 + 10x – 14y + 49 = 0
General equation of the circle: x2 + y2 + 2gx + 2fy + c = 0
Centre = (-g , -f ) and radius r = √( g2 + f2 – c)
Part – A
Example : Find the centre and radius of the circle x2 + y2 + 2x + 2y – 7 = 0 .
Soln: Given x2 + y2 + 2x + 2y – 7 = 0 .
We know the equation of circle is x2 + y2 + 2gx + 2fy + c = 0
2g = 2 2f = 2 c = -7
g = 1 f = 1
centre = ( -g , -f ) r = √( g2 + f2 – c)
= ( – 1 , – 1 ) r = √( (1)2 + (1)2 + 7)
r = √9 r = 3
| centre = (- 1 , – 1 ) & r = 3 |
Part – B
Example : Find the equation of the circle passing through the point A (2, – 3) and having its centre at
C ( – 5 , 1).
Soln: r = √(( – 5 – 2 )2 + (1 + 3 )2)
= √( (- 7)2 + (4)2 )
= √ ( 49 + 16)
r = √65
We know that the equation of circle is (x – h )2 + (y – k )2 = r2
Here h = -5, k = 1 (given) and r = √65
(x + 5 )2 + (y – 1 )2 = r2
x2 + 10x + 25 + y2 – 2y + 1 = 65
x2 + y2 + 10x – 2y + 26 – 65 = 0
x2 + y2 + 10x – 2y – 39 = 0
Therefore the equation of circle is x2 + y2 + 10x – 2y – 39 = 0
Equation of the Tangent to a circle at the point (x1,y1) on the circle
Equation of the tangent to a circle x2 + y2 + 2gx + 2fy + c = 0 at ( x1, y1) is
x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0
Part – B
Example : Find the equation of the tangent at (4,1) to the circle
x2 + y2 – 8x – 6y + 21 = 0 .
Soln: W.K.T the equation of tangent at ( x1, y1) is
x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0
Given x2 + y2 – 8x – 6y + 21 = 0
We know the equation of circle is x2 + y2 + 2gx + 2fy + c = 0
2g = – 8 2f = – 6 c = 21
g = – 4 f = – 3
x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0
x (4) + y ( 1) – 4 ( x + 4) – 3 ( y + 1) + 21 = 0
4x + y – 4x – 16 – 3y – 3 + 21 = 0
– 2y + 2 = 0
∴ Equation of tangent is y – 1 = 0.
FAMILY OF CIRCLES
Concentric Circles:
Two or more circles having the same centre but differ in radii are called concentric circles.
Note: Equation of concentric circles differ only by the constant term.
Example 1 :Show that the circles x2 + y2 – 4x + 2y +5 = 0 and x2 + y2 – 4x + 2y – 8 = 0
are concentric circles.
Soln: From the two given equations of the circles, we observe that the constant term alone differs
∴The given circles are concentric circles.
Part – A
Example 2 : Find the equation of the circle passing through the point (5 , 4 ) and concentric to the circle
x2 + y2 – 8x + 12y + 15 = 0 .
Soln: Equation of concentric circle be x2 + y2 – 8x + 12y + k = 0 ………….. ( 1 )
Put x = 5, y = 4 in ( 1 )
(5)2 + (4)2 – 8 ( 5 ) + 12 ( 4 ) + k = 0.
25 + 16 – 40 + 48 + k = 0.
89 – 40 + k = 0.
49 + k = 0
K = – 49
The required equation of the circle is x2 + y2 – 8x + 12y – 49 = 0
Orthogonal Circles:
Two circles x2 + y2 + 2g1x +2f1y+c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 are said to be Orthogonal Circles if
2g1g2 + 2 f1 f2 = c1 + c2
Part – B
Example :Prove that the circles x2 + y2 – 4x + 6y + 4 = 0 and
x2 + y2 + 2x + 4y + 4 = 0 cut orthogonally.
Soln: Given x2 + y2 – 4x + 6y + 4 = 0 ––––– (1) and
x2 + y2 + 2x + 4y + 4 = 0 ––––– (2)
From ( 1 )
2 g1 = – 4 2 f1 = 6 c1 = 4
g1= – 2 f1= 3
From ( 2 )
2 g2 =2 2 f2 = 4 c2 = 4
g2= 1 f2= 2
The condition for orthogonally is
2g1g2 + 2 f1 f2 = c1 + c2
2(-2)(1) + 2(3)( 4+ 4
-4 + 12 = 8
8 = 8
∴ The circles cut orthogonally.
Contact of Circles:
Case ( i ) :
Two circles touch externally if the distance between their centres is equal to sum of their radii.
i.e C1C2 = r1 + r2
Co-ordinates of point of contact are
P = ( (r1 x2 + r2 x1 ) / (r1 + r2 ) , (r1 y2 + r2 y1 ) / (r1 + r2 ) )
where C1 = (x1, y1) and C2 = (x2, y2)
Case ( ii ) :
Two circles touch internally if the distance between their centres is equal to difference of their radii.
i.e C1C2 = r1 – r2 or r2 – r1
Co-ordinates of point of contact are
P = ( (r1 x2 – r2 x1 ) / (r1 – r2 ) , (r1 y2 – r2 y1 ) / (r1 – r2 ) )
where C1 = (x1, y1) and C2 = (x2, y2)
Part – C
1. Prove that the circles x2 + y2 + 2x – 4y – 3 = 0 and x2 + y2 – 8x + 6y +7 = 0 touch each other.
Soln: Given x2 + y2 + 2x – 4y – 3 = 0 ––––– (1) and
x2 + y2 – 8x + 6y +7 = 0 ––––– (2)
From ( 1 )
2g1 = 2 2f1 = – 4 c1 = -3
g1 = 1 f1= – 2
centre is C1 = ( – g1 , – f1 ) r1 = √( g12 + f12 – c)
= ( -1 , 2 ) r1 = √( (1)2 + (-2)2 + 3)
r1 = √( 1 + 4 +3 )
r1 = √8 = 2√2
| C1= (-1 , 2) & r1 = 2√2 |
From ( 2 )
2g2 = -8 2f2 = 6 c2 = 7
g2 = – 4 f2 = 3
centre is C2 = ( – g2 , – f2 ) r2 = √( g22 + f22 – c)
= ( 4 , -3 ) r2 = √( (-4)2 + (3)2 – 7)
r2 = √( 16 + 9 – 7 )
r2 = √18 = √(9 × 2) = 3√2
| C2= ( 4 , -3 ) & r2 = 3√2 |
C1C2 = √(( – 1 – 4 )2 + (2 + 3 )2)
= √( (- 5)2 + (5)2 )
= √( 25 + 25)
= √50 = √(25 × 2) = 5√2
C1C2 = 5√2
r1 + r2 = 2√2 + 3√2
= 5√2 = C1C2
∴The circles touch each other externally.
2. Prove that the circles x2 + y2 – 4x + 6y – 112= 0 and x2 + y2 – 10x – 6y + 14 = 0 touch each other.
Soln: Given x2 + y2 – 4x + 6y – 112 = 0 ––––– (1) and
x2 + y2 – 10x – 6y + 14 = 0 ––––– (2)
From ( 1 )
2g1 = -4 2f1 = 6 c1 = -112
g1= -2 f1= 3
centre is C1 = ( – g1 , – f1 ) r1 = √( g12 + f12 – c)
= ( 2 , -3 ) r1 = √( (-2)2 + (3)2 + 112)
r1 = √( 4 + 9 + 112 )
r1 = √125 = √(125 × 5) = 5√5
| C1= ( 2 , -3 ) & r1 = 5√5 |
From ( 2 )
2 g2 = – 10 2 f2 = -6 c2 = 14
g2 = -5 f2 = – 3
centre is C2 = ( – g2 , – f2 ) r2 = √( g22 + f22 – c)
= ( 5 , 3 ) r2 = √( (-5)2 + (-3)2 – 14)
r2 = √( 25 + 9 – 14 )
r2 = √20 = √(4 × 5)= 2√5
| C2= ( 5 , 3 ) & r2 = 2√5 |
C1 C2 = √(( 5 – 2 )2 + (3 + 3 )2)
= √( (3)2 + (6)2 )
= √( 9 + 36)
C1 C2 = √45 = √(9× 5) = 3√5
r1 – r2 = 5√5 –
= 3√5 = C1 C2
∴The circles touch each other internally.
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