1.2 ANALYTICAL GEOMETRY II

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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.

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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

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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)

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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  

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  

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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

Tangent to a Circle: Formulas, Properties, Theorems

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

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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.

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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. 

√ Concentric Circles (Definition and Example) | Σ Tricks
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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

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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

How to Show Two Circles are Orthogonal

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)(2) =      4+ 4

                            -4   + 12        =      8

                          8                      =        8

∴ The circles cut orthogonally.

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Contact of Circles:

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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 ) :

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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)

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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.

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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  –   2√5

                      =  3√5  =  C1 C2

∴The circles touch each other internally.

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