The points A (2,4) and B ( 6,0) lie on a circle.

Question

B) The points A (2,4) and B ( 6,0) lie on a circle.

B I)Find the midpoint of AB and hence the equation of a straight line on which the centre of the circle lies. 

B II) If x = -1 at the centre of the circle, write down the equation of the circle.

Answer 1

(B I).

The two points are A(2, 4) and B(6, 0).

Substitute A = (x₁, y₁) = (2, 4) and B = (x₂, y₂) = (6, 0) in the midpoint formula: M = [(x₁ + x₂)/2, (y₁ + y₂)/2]

M = [(2 + 6)/2, (4 + 0)/2] = (8/2, 4/2) = (4, 2).

Therefore, center of the circle: (h, k) = M = (4, 2).

Substitute A = (x₁, y₁) = (2, 4) and B = (x₂, y₂) = (6, 0) in the distance formula: AB = √[(x₂ - x₁)² + (y₂ - y₁)²]

AB = √[(6 - 2)² + (0 - 2)²] = √[4² + (-2)²] = √[4² + 4] = √20.

Radius of the circle = r = AB = √20.

Substitute (h, k) = (4, 2) and r² = 20 in the standard form of circle equation: (x - h)² + (y - k)² = r².

(x - 4)² + (y - 2)² = 20.

Answer 2

(B II).The standard form of circle equation is (x - h)² + (y - k)² = r²

Where (h, k) is center and r is radius.

Substitute for (x, y) = (2, 4) and h = - 1 in standard form.

[2 - (- 1)]² + [4 - k]² = r²

9 + 16 + k² - 8k = r²

25 + k² - 8k = r² ---> (1)

Substitute for (x, y) = (6, 0) and h = - 1 in standard form.

[6 - (- 1)]² + [0 - k]² = r²

49 + k²  = r² - --> (2)

Equate the equations (1) and (2).

25 + k² - 8k = 49 + k²

25 - 8k = 49

8k = 25 - 49

8k = - 24

k = - 3

Radius is distance between center and any point lies on the circle.

Find the distance between ( -1, - 3) and (2, 4)

(x₁, y₁) = (- 1, - 3) and B = (x₂, y₂) = (2, 4)

= √[(x₂ - x₁)² + (y₂ - y₁)²

= √[(2 + 1)² + (4 + 3)²

= √9+49

= √58

Substitute for (h, k) = (- 1, - 3) and r = √58 in standard form.

Equation of circle [(x - ( -1))² + [y - (- 3)]² = (√58)²

Required equation is (x + 1)² + (y + 3)² = 58.