Question

Show that the points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle, and find its equation, centre, and radius.

Answer 1

Given:

The points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle.

Let us assume the circle passes through the points A, B, C.

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

x² + y² + 2ax + 2by + c = 0….. (1)

Substituting A (5, 5) in (1), we get,

5² + 5² + 2a(5) + 2b(5) + c = 0

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

10a + 10b + c + 50 = 0….. (2)

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

6² + 4² + 2a(6) + 2b(4) + c = 0

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

12a + 8b + c + 52 = 0….. (3)

Substitute the point C (-2, 4) in equation (1), we get,

(-2)² + 4² + 2a(-2) + 2b(4) + c = 0

4 + 16 – 4a + 8b + c = 0

20 – 4a + 8b + c = 0

4a – 8b – c – 20 = 0….. (4)

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

a = – 2, b = – 1 and c = – 20

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

x² + y² + 2(- 2)x + 2(- 1)y – 20 = 0

x² + y² – 4x – 2y – 20 = 0 ….. (5)

Substituting D (7, 1) in equation (5) we get,

7² + 1² – 4(7) – 2(1) – 20

49 + 1 – 28 – 2 – 20

0

∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) lie on a circle.

Now let us find the centre and the radius.
We know that for a circle x² + y² + 2ax + 2by + c = 0,

Centre = (-a, -b)

Radius = √(a² + b² – c) 

Comparing equation (5) with equation (1), we get

Centre = [-(-4)/2, -(-2)/2)]

= (2, 1)

Radius = √(2² + 1² – (-20)) 

= √(25)

= 5

∴ The centre and radius of the circle is (2, 1) and 5.