Question

Find the equation of the circle passing through the points :
(i) (5, 7), (8, 1) and (1, 3)

(ii) (1, 2), (3, – 4) and (5, – 6)

(iii) (5, -8), (-2, 9) and (2, 1)

(iv) (0, 0), (-2, 1) and (-3, 2)

Answer 1

(i) (5, 7), (8, 1) and (1, 3)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

Firstly let us find the values of a, b and c

Substitute the given point (5, 7) in equation (1), we get

5² + 7² + 2a (5) + 2b (7) + c = 0

25 + 49 + 10a + 14b + c = 0

10a + 14b + c + 74 = 0….. (2)

By substituting the given point (8, 1) in equation (1), we get

8² + 1² + 2a (8) + 2b (1) + c = 0

64 + 1 + 16a + 2b + c = 0

16a + 2b + c + 65 = 0….. (3)

Substituting the point (1, 3) in equation (1), we get

1² + 3² + 2a (1) + 2b (3) + c = 0

1 + 9 + 2a + 6b + c = 0

2a + 6b + c + 10 = 0….. (4)

Now by simplifying the equations (2), (3), (4) we get the values

a = -29/6, b = -19/6, c = 56/3

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

x² + y² + 2 (-29/6)x + 2 (-19/6) + 56/3 = 0

x² + y² – 29x/3 – 19y/3 + 56/3 = 0

3x² + 3y² – 29x – 19y + 56 = 0

∴ The equation of the circle is 3x² + 3y² – 29x – 19y + 56 = 0

(ii) (1, 2), (3, – 4) and (5, – 6)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

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

1² + 2² + 2a (1) + 2b (2) + c = 0

1 + 4 + 2a + 4b + c = 0

2a + 4b + c + 5 = 0….. (2)

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

3² + (- 4)² + 2a (3) + 2b (- 4) + c = 0

9 + 16 + 6a – 8b + c = 0

6a – 8b + c + 25 = 0….. (3)

Substitute the points (5, -6) in equation (1), we get

5² + (- 6)² + 2a (5) + 2b (- 6) + c = 0

25 + 36 + 10a – 12b + c = 0

10a – 12b + c + 61 = 0….. (4)

Now by simplifying the equations (2), (3), (4) we get

a = – 11, b = – 2, c = 25

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

x² + y² + 2(- 11)x + 2(- 2) + 25 = 0

x² + y² – 22x – 4y + 25 = 0

∴ The equation of the circle is x² + y² – 22x – 4y + 25 = 0

(iii) (5, -8), (-2, 9) and (2, 1)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

Substitute the point (5, -8) in equation (1), we get

5² + (- 8)² + 2a(5) + 2b(- 8) + c = 0

25 + 64 + 10a – 16b + c = 0

10a – 16b + c + 89 = 0….. (2)

Substitute the points (-2, 9) in equation (1), we get

(- 2)² + 9² + 2a(- 2) + 2b(9) + c = 0

4 + 81 – 4a + 18b + c = 0

-4a + 18b + c + 85 = 0….. (3)

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

2² + 1² + 2a(2) + 2b(1) + c = 0

4 + 1 + 4a + 2b + c = 0

4a + 2b + c + 5 = 0….. (4)

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

a = 58, b = 24, c = – 285.

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

x² + y² + 2(58)x + 2(24) – 285 = 0

x² + y² + 116x + 48y – 285 = 0

∴ The equation of the circle is x² + y² + 116x + 48y – 285 = 0

(iv) (0, 0), (-2, 1) and (-3, 2)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

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….. (2)

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

(- 2)² + 1² + 2a(- 2) + 2b(1) + c = 0

4 + 1 – 4a + 2b + c = 0

-4a + 2b + c + 5 = 0….. (3)

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

(- 3)² + 2² + 2a(- 3) + 2b(2) + c = 0

9 + 4 – 6a + 4b + c = 0

-6a + 4b + c + 13 = 0….. (4)

By simplifying the equations (2), (3), (4) we get

a = -3/2, b = -11/2, c = 0

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

x² + y² + 2(-3/2)x + 2(-11/2)y + 0 = 0

x² + y² – 3x – 11y = 0

∴ The equation of the circle is x² + y² – 3x – 11y = 0