# 1.2 ANALYTICAL GEOMETRY II

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  x+ y2  = r2

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#### 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 = 52

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)  https://clnk.in/qagW

#### 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

Example  :   Find the centre and radius of the circle  x2  +   y2 – 8y  + 3 = 0 .

Soln:    Given    x2  +   y2 – 8y  + 3 = 0 .

We know the equation of circle is  x2  +   y2  + 2gx  + 2fy  + c = 0

2g  = 0                  2f = -8                  c = 3

g  =  0                      f =  -4

centre = ( -g , -f )                              r = √( g2  + f2 – c)

=   ( 0 , 4 )                           r = √( (0)2  + (- 4)2 – 3)

r = √13

centre =  (- 1 , – 1 )      &        r =  3 https://clnk.in/qfGm

#### 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 =   (√65) 2

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 https://clnk.in/qfGp

#### 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. https://clnk.in/qfxR

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 https://clnk.in/qfGy

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 + c

2(-2)(1) + 2(3)(2) =      4+ 4

-4   + 12        =      8

8                      =        8

∴ The circles cut orthogonally.

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### Contact of Circles:

Case ( i ) :

Two circles touch externally if the distance between their centres is equal to sum of their radii.

i.e  C1C=  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  C1C= r1  –  ror   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) https://clnk.in/qfGF

#### Part – C

1.   Prove that the circles  x2  +   y2  + 2x  –  4y  – 3 = 0 and x2  +   y– 8x  +  6y  +7 = 0 touch each  other.

Soln:    Given  x2  +   y2  + 2x  –  4y  – 3 = 0 ––––– (1)      and

x2  +   y– 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 )                 r= √( (1)2  + (-2)2 + 3)

r= √( 1  + 4  +3 )

r1 = √8 = 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 )                  r= √( (-4)2  + (3)2  – 7)

r= √( 16  + 9  – 7 )

r2 = √18 =   √(9 × 2)  =   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 )                        r= √( (-2)2  + (3)2 + 112)

r= √( 4  + 9  + 112 )

r1 = √125  = √(125 × 5)  =   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 )                   r= √( (-5)2  + (-3)2  – 14)

r= √( 25  + 9  – 14 )

r2 = √20  = √(4 × 5)= 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  –   2√5

=  3√5  =  C1 C2

∴The circles touch each other internally.

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