Equation of Circle (Centre Radius Form)

Type – I: Equation of Circle When Centre and Radius are Given

ALGORITHM (Steps):

  1. Write given centre ≡ (h, k) and given radius = r
  2. Use centre radius form (x – h)2 + (y – k)2 = r2
  3. Simplify the equation and write it in standard form ax2 + by2 + 2gx + 2fy + c = 0

Example – 01:

  • Find equation of circle having centre  at (2, -3) and radius 5.
  • Solution:

Given centre (2, -3) ≡ (h, k) and radius = r = 5

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x – 2)2 + (y + 3)2 = 52



∴  x2 – 4x + 4 + y2 + 6y + 9 = 25

∴  x2 + y– 4x + 6y + 13 – 25 = 0

∴  x2 + y– 4x + 6y – 12 = 0

Ans: The required equation of circle is x2 + y– 4x + 6y – 12 = 0



Example – 02:

  • Find equation of circle having centre  at origin and radius 4.
  • Solution:

Given centre (0, 0) ≡ (h, k) and radius = r = 4

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x – 0)2 + (y + 0)2 = 42

∴  x2  + y= 16



Ans: The required equation of circle is x2 + y = 16

Example – 03:

  • Find equation of circle having centre  at (3, -2) and radius 5.
  • Solution:

Given centre (3, -2) ≡ (h, k) and radius = r = 5

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x – 3)2 + (y + 2)2 = 52



∴  x2 – 6x + 9 + y2 + 4y + 4 = 25

∴  x2 + y– 6x + 4y + 13 – 25 = 0

∴  x2 + y– 6x + 4y – 12 = 0

Ans: The required equation of circle is x2 + y– 6x + 4y – 12 = 0

Example – 04:

  • Find equation of circle having centre  at (-3, -2) and radius 6.
  • Solution:

Given centre (-3, -2) ≡ (h, k) and radius = r = 6



By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x + 3)2 + (y + 2)2 = 62

∴  x2 + 6x + 9 + y2 + 4y + 4 = 36



∴  x2 + y+ 6x + 4y + 13 – 36 = 0

∴  x2 + y+ 6x + 4y – 23 = 0

Ans: The required equation of circle is x2 + y+ 6x + 4y  – 23 = 0

Example – 05:

  • Find equation of circle having centre  at (a cosα, a sinα) and radius a.
  • Solution:

Given centre (a cosα, a sinα) ≡ (h, k) and radius = r = 6

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2



(x – a cosα)2 + (y – a sinα)2 = a2

∴  x2 – 2xa cosα + a2 cos2α + y2 – 2ya sinα + a2 sin2α = a2

∴  x2 + y2  – 2ax cosα  – 2ay sinα + a2 cos2α  + a2 sin2α = a2

∴  x2 + y2  – 2ax cosα  – 2ay sinα + a2 (cos2α  + sin2α) – a2  = 0



∴  x2 + y2  – 2ax cosα  – 2ay sinα + a2 (1) – a2  = 0

∴  x2 + y2  – 2acosα.x  – 2a sinα.y = 0

Ans: The required equation of circle is x2 + y2  – 2acosα.x  – 2a sinα.y = 0

Example – 06:

  • Find equation of circle having centre  at (a, b) and radius  √a2+b2
  • Solution:

Given centre (a, b) ≡ (h, k) and radius = r = √a2+b

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2



∴  (x – a)2 + (y – b)2 = (√a2+b )2

∴  x2 – 2ax + a2 + y2 – 2by + b2 = a2 + b2

∴  x2 + y– 2ax – 2by + a2  + b2 – a2 – b2 = 0

∴  x2 + y– 2ax – 2by  = 0

Ans: The required equation of circle is x2 + y– 2ax – 2by  = 0

Type – II: Equation of Circle When Centre and Point on the Circle are Given

ALGORITHM (Steps):

centre radius form

  1. Write given centre C ≡ (h, k) and given point say P (x, y)
  2. Use distance formula to find CP represent it as radius = r
  3. Use centre radius form (x – h)2 + (y – k)2 = r2
  4. Simplify the equation and write it in standard form ax2 + by2 + 2gx + 2fy + c = 0

Example – 07:

  • Find equation of circle having centre  at (-3, 1) and passing through (5, 2)
  • Solution:

Given centre C(-3, 1) ≡ (h, k) and  point on circle P(5, 2)

By distance formula

CP2 = (5 + 3)2 + (2 – 1)2 = 64 + 1 = 65

∴  CP = √65

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x + 3)2 + (y – 1)2 = (√65)2

∴  x2 + 6x + 9 + y2 – 2y + 1 = 65

∴  x2 + y+ 6x – 2y + 10 – 65 = 0

∴  x2 + y+ 6x – 2y – 55 = 0

Ans: The required equation of circle is x2 + y+ 6x – 2y – 55 = 0

Example – 08:

  • Find equation of circle having centre  at (2, -1) and passing through (3, 6)
  • Solution:

Given centre C(2, -1) ≡ (h, k) and  point on circle P(3, 6)

By distance formula

CP2 = (3 – 2)2 + (6 + 1)2 =1 + 49 = 50

∴  CP = √50

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x – 2)2 + (y + 1)2 = (√50)2

∴  x2 – 4x + 4 + y2 + 2y + 1 = 50

∴  x2 + y– 4x + 2y + 5 – 50 = 0

∴  x2 + y– 4x + 2y – 45 = 0

Ans: The required equation of circle is x2 + y– 4x + 2y – 45 = 0

Example – 09:

  • Find equation of circle having centre  at (2, -3) and passing through P(-3, 5)
  • Solution:

Given centre C(2, – 3) ≡ (h, k) and  point on circle (-3, 5)

By distance formula

CP2 = (-3 – 2)2 + (5 + 3)2 = 25 + 68 = 89

∴  CP = √89

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x – 2)2 + (y + 3)2 = (√89)2

∴  x2 – 4x + 4 + y2 + 6y + 9 = 89

∴  x2 + y– 4x + 6y + 13 – 89 = 0

∴  x2 + y– 4x + 6y – 76 = 0

Ans: The required equation of circle is x2 + y– 4x + 6y – 76 = 0

Example – 10:

  • Find equation of circle passing through (-3, 2) and having centre at the point of intersection of lines x – 2y = 4 and 2x + 5y + 1 = 0
  • Solution:

Solving the equations x – 2y = 4 and 2x + 5y + 1 = 0 we get x = 2 and y = -1

Thus centre C(2, – 1) ≡ (h, k) and  point on circle P(-3, 2)

By distance formula

CP2 = (-3 – 2)2 + (2 + 1)2 = 25 + 9 = 34

∴  CP = √34

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x – 2)2 + (y + 1)2 = (√34)2

∴  x2 – 4x + 4 + y2 + 2y + 1 = 34

∴  x2 + y– 4x + 2y + 5 – 34 = 0

∴  x2 + y– 4x + 2y – 29 = 0

Ans: The required equation of circle is x2 + y– 4x + 2y – 29 = 0

Example – 11:

  • Find equation of circle passing through intersection of x + 3y = 0 and 2x – 7y = 0 and having centre at the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0
  • Solution:

Solving the equations x + 3y = 0 and 2x – 7y = 0 we get x = 0 and y = 0

Hence the circle passes through O(0, 0)

Solving the equations x + y + 1 = 0 and x – 2y + 4 = 0 we get x = -2 and y = 1

Hence the centre of circle is C(-2, 1)

Thus centre C(-2, 1) ≡ (h, k) and  point on circle O(0, 0)

By distance formula

CO2 = (0 + 2)2 + (0 – 1)2 = 4 + 1 = 5

∴  CO = √5

By centre radius form, the equation of circle is given by

(x – h)2 + (y – k)2 = r2

∴  (x + 2)2 + (y – 1)2 = (√5)2

∴  x2 + 4x + 4 + y2 – 2y + 1 = 5

∴  x2 + y+ 4x – 2y + 5 – 5 = 0

∴  x2 + y+ 4x – 2y = 0

Ans: The required equation of circle is x2 + y+ 4x – 2y = 0

 

 

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