(i) x + y + 3 = 0, x – y + 1 = 0 and x = 3

Given:

The lines x + y + 3 = 0

x – y + 1 = 0

x = 3

On solving these lines we get the intersection points A (-2, -1), B (3, 4), C (3, -6)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (-2, -1) in equation (1), we get

(- 2)² + (- 1)² + 2a(-2) + 2b(-1) + c = 0

4 + 1 – 4a – 2b + c = 0

5 – 4a – 2b + c = 0

4a + 2b – c – 5 = 0….. (2)

Substitute the points (3, 4) in equation (1), we get

3² + 4² + 2a(3) + 2b(4) + c = 0

9 + 16 + 6a + 8b + c = 0

6a + 8b + c + 25 = 0….. (3)

Substitute the points (3, -6) in equation (1), we get

3² + (- 6)² + 2a(3) + 2b(- 6) + c = 0

9 + 36 + 6a – 12b + c = 0

6a – 12b + c + 45 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = – 3, b = 1, c = -15.

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 3)x + 2(1)y – 15 = 0

x² + y² – 6x + 2y – 15 = 0

∴ The equation of the circle is x² + y² – 6x + 2y – 15 = 0.

(ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0

Given:

The lines 2x + y – 3 = 0

x + y – 1 = 0

3x + 2y – 5 = 0

On solving these lines we get the intersection points A(2, – 1), B(3, – 2), C(1,1)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (2, -1) in equation (1), we get

2² + (- 1)² + 2a(2) + 2b(- 1) + c = 0

4 + 1 + 4a – 2b + c = 0

4a – 2b + c + 5 = 0….. (2)

Substitute the points (3, -2) in equation (1), we get

3² + (- 2)² + 2a(3) + 2b(- 2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (3)

Substitute the points (1, 1) in equation (1), we get

1² + 1² + 2a(1) + 2b(1) + c = 0

1 + 1 + 2a + 2b + c = 0

2a + 2b + c + 2 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = -13/2, b = -5/2, c = 16

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2 (-13/2)x + 2 (-5/2)y + 16 = 0

x² + y² – 13x – 5y + 16 = 0

∴ The equation of the circle is x² + y² – 13x – 5y + 16 = 0

(iii) x + y = 2, 3x – 4y = 6 and x – y = 0

Given:

The lines x + y = 2

3x – 4y = 6

x – y = 0

On solving these lines we get the intersection points A(2,0), B(- 6, – 6), C(1,1)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (2, 0) in equation (1), we get

2² + 0² + 2a(2) + 2b(0) + c = 0

4 + 4a + c = 0

4a + c + 4 = 0….. (2)

Substitute the point (-6, -6) in equation (1), we get

(- 6)² + (- 6)² + 2a(- 6) + 2b(- 6) + c = 0

36 + 36 – 12a – 12b + c = 0

12a + 12b – c – 72 = 0….. (3)

Substitute the points (1, 1) in equation (1), we get

1² + 1² + 2a(1) + 2b(1) + c = 0

1 + 1 + 2a + 2b + c = 0

2a + 2b + c + 2 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = 2, b = 3, c = – 12.

Substituting the values of a, b, c in equation (1), we get

x² + y² + 2(2)x + 2(3)y – 12 = 0

x² + y² + 4x + 6y – 12 = 0

∴ The equation of the circle is x² + y² + 4x + 6y – 12 = 0

(iv) y = x + 2, 3y = 4x and 2y = 3x

Given:

The lines y = x + 2

3y = 4x

2y = 3x

On solving these lines we get the intersection points A(6,8), B(0,0), C(4,6)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (6, 8) in equation (1), we get

6² + 8² + 2a(6) + 2b(8) + c = 0

36 + 64 + 12a + 16b + c = 0

12a + 16b + c + 100 = 0…… (2)

Substitute the points (0, 0) in equation (1), we get

0² + 0² + 2a(0) + 2b(0) + c = 0

0 + 0 + 0a + 0b + c = 0

c = 0….. (3)

Substitute the points (4, 6) in equation (1), we get

4² + 6² + 2a(4) + 2b(6) + c = 0

16 + 36 + 8a + 12b + c = 0

8a + 12b + c + 52 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = – 23, b = 11, c = 0

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 23)x + 2(11)y + 0 = 0

x² + y² – 46x + 22y = 0

∴ The equation of the circle is x² + y² – 46x + 22y = 0