# Circles and midpoitnts

A)Find the points of intersection of the line y= 2x +1 and the circle (x-7)^2 + y^2 = 125

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.

If you can, please show steps so i can learn!
asked Nov 3, 2014 in CALCULUS

+1 vote

(A).

The line: y = 2x + 1 and the circle: (x-7)2 + y2 = 125.

Substitute y = 2x + 1 in the circle equation.

(x-7)2 + (2x+1)2 = 125

x2 - 14x + 49 + 4x2 + 1 + 4x = 125

5x2 - 10x + 50 = 125

x2 - 2x + 10 = 25

x2 - 2x - 15 = 0

x2 - 5x + 3x - 15 = 0

x(x - 5) + 3(x - 5) = 0

(x + 3)(x - 5) = 0

x + 3 = 0 and x - 5 = 0

x = -3 and x = 5.

If x = -3 ⇒ y = 2(-3) + 1 = -5

If x = 5 ⇒ y = 2(5) + 1 = 11.

The intersection points are (-3, -5) and (5, 11).

+1 vote

(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.

(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.