1. A(-3;4) is a point on a circle

with centre at the origin.
a.determine the coordinates of B, if AB is a diameter.
b. Show that point C(0;5) lies on this circle.
c.show that ACB has a magnitude of 90degrees.

2.a.ddetermine the equation of the circle with AC as diameter, if A is the point (-5;12) and C(5;-12).
b.if B is the point (0;13) and D(-7;-2#30) show that ABCD is a cyclic quadrilateral, that all the vertices lie on the

#1

clearly the circle is of radius 5. So, since x^2+y^2 = 5^2, (0,5) lies on the circle.

the center is the midpoint of AC: (0,0)
so, the radius is 13.

0^2+13^2 = 13^2
7^2 + (2√30)^2 = 49+120 = 13^2

so, all vertices lie on the circle.

To answer these questions, we will use the distance formula, which is derived from the Pythagorean theorem, and the concept of perpendicular lines.

1. a. To determine the coordinates of point B, if AB is a diameter of the circle with center at the origin, we need to find the midpoint of the line segment AB. The midpoint formula states that the x-coordinate of the midpoint is the average of the x-coordinates of the endpoints, and the y-coordinate of the midpoint is the average of the y-coordinates of the endpoints.

Given point A (-3, 4), and since the center of the circle is at the origin (0, 0), the x-coordinate of the midpoint is (0 + (-3))/2 = -1.5, and the y-coordinate of the midpoint is (4 + 0)/2 = 2.

Therefore, the coordinates of point B are (-1.5, 2).

b. To show that point C (0, 5) lies on this circle, we need to calculate the distance between the center of the circle (0, 0) and point C (0, 5). If this distance is equal to the radius, then point C lies on the circle.

Using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((0 - 0)^2 + (5 - 0)^2)
Distance = sqrt(0 + 25)
Distance = sqrt(25)
Distance = 5

Since the distance between the center of the circle and point C is 5, which is equal to the radius, point C lies on the circle.

c. To show that triangle ACB has a magnitude of 90 degrees, we need to show that the line segment AB is perpendicular to the line segment AC.

To determine if two lines are perpendicular, we need to check if the product of their slopes is -1. The slope of a line is calculated using the formula: slope = (change in y) / (change in x).

The slope of line AB = (2 - 4) / (-1.5 - (-3)) = -2 / (1.5) = -4/3
The slope of line AC = (5 - 0) / (0 - (-3)) = 5 / 3

The product of the slopes of AB and AC is (-4/3) * (5/3) = -20/9

Since the product of the slopes is not -1, we conclude that triangle ACB does not have a magnitude of 90 degrees.

2. a. To determine the equation of the circle with AC as the diameter, we need to find the center and radius of the circle. The center of the circle is the midpoint of the line segment AC, and the radius is half the length of AC.

Given point A (-5, 12) and point C (5, -12), we can find the midpoint of AC:
Midpoint_x = (-5 + 5) / 2 = 0
Midpoint_y = (12 + (-12)) / 2 = 0

Therefore, the center of the circle is (0, 0).

To find the radius, we calculate half the length of AC:
Radius = sqrt((5 - (-5))^2 + ((-12) - 12)^2) / 2
Radius = sqrt(10^2 + (-24)^2) / 2
Radius = sqrt(100 + 576) / 2
Radius = sqrt(676) / 2
Radius = sqrt(169)
Radius = 13/2

The equation of the circle with center (0, 0) and radius 13/2 is:
(x - 0)^2 + (y - 0)^2 = (13/2)^2
x^2 + y^2 = 169/4

b. To show that points A, B, C, and D form a cyclic quadrilateral, we need to prove that the opposite angles of the quadrilateral are supplementary, or their sum is 180 degrees.

Let's calculate the slopes of the sides AB, BC, CD, and DA, and check if opposite sides are perpendicular using the product of slopes method explained in question 1.

Slope of AB = (13 - 12) / (0 - (-5)) = 1/5
Slope of BC = (13 - (-12)) / (0 - 5) = 25/5 = 5
Slope of CD = ((-2#30) - (-12)) / (-7 - 5) = (-2#30 + 12) / (-12) = -2#30 / 12 = -5/3
Slope of DA = ((-2#30) - 12) / (-7 - (-5)) = (-2#30 - 12) / (-2) = -2#30 / 2 = -#30

The product of slopes for opposite sides:
Product of AB and CD = (1/5) * (-5/3) = -1
Product of BC and DA = (5) * (-#30) = -1

Since the product of slopes for opposite sides is -1, we can conclude that opposite sides are perpendicular. Therefore, quadrilateral ABCD is a cyclic quadrilateral, meaning all of its vertices lie on a circle.