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

#### ALGORITHM (Steps):

- Write given centre ≡ (h, k) and given radius = r
- Use centre radius form (x – h)
^{2}+ (y – k)^{2}= r^{2} - Simplify the equation and write it in standard form ax
^{2}+ by^{2}+ 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} = r^{2}

∴ (x – 2)^{2} + (y + 3)^{2} = 5^{2}

∴ x^{2} – 4x + 4 + y^{2} + 6y + 9 = 25

∴ x^{2} + y^{2 }– 4x + 6y + 13 – 25 = 0

∴ x^{2} + y^{2 }– 4x + 6y – 12 = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }– 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} = r^{2}

∴ (x – 0)^{2} + (y + 0)^{2} = 4^{2}

∴ x^{2} + y^{2 }= 16

**Ans:** The required equation of circle is x^{2} + y^{2 } = 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} = r^{2}

∴ (x – 3)^{2} + (y + 2)^{2} = 5^{2}

∴ x^{2} – 6x + 9 + y^{2} + 4y + 4 = 25

∴ x^{2} + y^{2 }– 6x + 4y + 13 – 25 = 0

∴ x^{2} + y^{2 }– 6x + 4y – 12 = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }– 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} = r^{2}

∴ (x + 3)^{2} + (y + 2)^{2} = 6^{2}

∴ x^{2} + 6x + 9 + y^{2} + 4y + 4 = 36

∴ x^{2} + y^{2 }+ 6x + 4y + 13 – 36 = 0

∴ x^{2} + y^{2 }+ 6x + 4y – 23 = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }+ 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} = r^{2}

(x – a cosα)^{2} + (y – a sinα)^{2} = a^{2}

∴ x^{2} – 2xa cosα + a^{2} cos^{2}α + y^{2} – 2ya sinα + a^{2} sin^{2}α = a^{2}

∴ x^{2} + y^{2 }– 2ax cosα – 2ay sinα + a^{2} cos^{2}α + a^{2} sin^{2}α = a^{2}

∴ x^{2} + y^{2 }– 2ax cosα – 2ay sinα + a^{2} (cos^{2}α + sin^{2}α) – a^{2 }= 0

∴ x^{2} + y^{2 }– 2ax cosα – 2ay sinα + a^{2} (1) – a^{2 }= 0

∴ x^{2} + y^{2 }– 2acosα.x – 2a sinα.y = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }– 2acosα.x – 2a sinα.y = 0

#### Example – 06:

- Find equation of circle having centre at (a, b) and radius √a
^{2}+b^{2} **Solution:**

Given centre (a, b) ≡ (h, k) and radius = r = √a^{2}+b^{2 }

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

(x – h)^{2} + (y – k)^{2} = r^{2}

∴ (x – a)^{2} + (y – b)^{2} = (√a^{2}+b^{2 } )^{2}

∴ x^{2} – 2ax + a^{2} + y^{2} – 2by + b^{2} = a^{2} + b^{2}

∴ x^{2} + y^{2 }– 2ax – 2by + a^{2} + b^{2} – a^{2} – b^{2} = 0

∴ x^{2} + y^{2 }– 2ax – 2by = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }– 2ax – 2by = 0

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

#### ALGORITHM (Steps):

- Write given centre C ≡ (h, k) and given point say P (x, y)
- Use distance formula to find CP represent it as radius = r
- Use centre radius form (x – h)
^{2}+ (y – k)^{2}= r^{2} - Simplify the equation and write it in standard form ax
^{2}+ by^{2}+ 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

CP^{2} = (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} = r^{2}

∴ (x + 3)^{2} + (y – 1)^{2} = (√65)^{2}

∴ x^{2} + 6x + 9 + y^{2} – 2y + 1 = 65

∴ x^{2} + y^{2 }+ 6x – 2y + 10 – 65 = 0

∴ x^{2} + y^{2 }+ 6x – 2y – 55 = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }+ 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

CP^{2} = (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} = r^{2}

∴ (x – 2)^{2} + (y + 1)^{2} = (√50)^{2}

∴ x^{2} – 4x + 4 + y^{2} + 2y + 1 = 50

∴ x^{2} + y^{2 }– 4x + 2y + 5 – 50 = 0

∴ x^{2} + y^{2 }– 4x + 2y – 45 = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }– 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

CP^{2} = (-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} = r^{2}

∴ (x – 2)^{2} + (y + 3)^{2} = (√89)^{2}

∴ x^{2} – 4x + 4 + y^{2} + 6y + 9 = 89

∴ x^{2} + y^{2 }– 4x + 6y + 13 – 89 = 0

∴ x^{2} + y^{2 }– 4x + 6y – 76 = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }– 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

CP^{2} = (-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} = r^{2}

∴ (x – 2)^{2} + (y + 1)^{2} = (√34)^{2}

∴ x^{2} – 4x + 4 + y^{2} + 2y + 1 = 34

∴ x^{2} + y^{2 }– 4x + 2y + 5 – 34 = 0

∴ x^{2} + y^{2 }– 4x + 2y – 29 = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }– 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

CO^{2} = (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} = r^{2}

∴ (x + 2)^{2} + (y – 1)^{2} = (√5)^{2}

∴ x^{2} + 4x + 4 + y^{2} – 2y + 1 = 5

∴ x^{2} + y^{2 }+ 4x – 2y + 5 – 5 = 0

∴ x^{2} + y^{2 }+ 4x – 2y = 0

**Ans:** The required equation of circle is x^{2} + y^{2 }+ 4x – 2y = 0