# Find the standard equation of the circle that contains endpoints of a diameter at (2,-1) and (-4, -3)

#1

Find the standard equation of the circle that contains endpoints of a diameter at (2,-1) and (-4, -3)

The general equation of a circle is: (x – h)^2 + (y – k)^2 = r^2

Finding the center (h,k):

Since the center is the midpoint of the line segment with endpoints (2,-1) and (-4,-3), we need to find the midpoint.

X-Coordinate of Midpoint = [x(1) + x(2)] / 2 = [2 + (-4)] / 2 = -2/2 = -1

Since the x coordinate of midpoint is -1, this means that h = -1

Y-Coordinate of Midpoint = [y(1) + (y(2)] / 2 = [-1 + (-3)] / 2 = -4/2 = -2

Since the y coordinate of midpoint is -2, this means that k = -2

So the center is the point (-1, -2)

Now let’s find the radius squared

Use the formula (x – h)^2 + (y – k)^2 = r^2, where (h,k) is the center and (x,y) is an arbitrary point on the circle.

In this case, x = 2, y = -1 and h = -1, k = -2. Plug these values into the equation above and
simplify to get:

[2 – (-1)]^2 + [(-1) – (-2)]^2 = r^2
r^2 = 3^2 + 1^2 = 9 + 1 = 10
r^2 = 10.

So because h = -1, k = -2, and r^2 = 10, this means that the equation of the circle that passes through the points (2,-1) and (-4,-3), (which are the endpoints of the diameter) is

[(x – (-1)] + [(y – (-2)] = 10

(x + 1)^2 + (y + 2)^2 = 10 ANSWER