Question

Find an equation of the circle whose diameter has endpoints ( −1,−4) and (3,2) .

Answer 1

The equation of the circle with diameter endpoints (-1,-4) and (3,2) is (x - 1)^2 + (y + 1)^2 = 13, found by calculating the center at (1, -1) and the radius as sqrt(52)/2.

Step-by-step explanation:

To find an equation of the circle whose diameter has endpoints at (-1,-4) and (3,2), we need to determine the center and radius of the circle. The center of a circle is the midpoint of the diameter, and the radius is half the length of the diameter.

First, we calculate the midpoint (which will be the center of the circle) using the formula: (x1 + x2)/2, (y1 + y2)/2. For the given points (-1,-4) and (3,2), the midpoint is ((-1 + 3)/2, (-4 + 2)/2), which simplifies to (1, -1). So, the center of the circle is (1, -1).

To find the radius, we use the distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2). The distance between the two points is sqrt((3 - (-1))^2 + (2 - (-4))^2), which simplifies to sqrt(4^2 + 6^2) = sqrt(16 + 36) = sqrt(52). The radius is half of the diameter, so r = sqrt(52)/2.

The standard form of a circle's equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius. Substituting the values we found, we get (x - 1)^2 + (y + 1)^2 = (sqrt(52)/2)^2, which simplifies to (x - 1)^2 + (y + 1)^2 = 13. This is the equation of the circle.

Answer 2

The equation of a circle is:

(x-h)^2+(y-k)^2=r^2, where (h,k) is the center and r is the radius

So the first thing we can do is to find the center of the circle, which will be at the midpoint of the two endpoints of the diameter. The midpoint is simply the average of the coordinates of the endpoints.

mp=((-1+3)/2, (-4+2)/2)

mp=(1, -1), and this midpoint is the center of the circle our (h,k) for the circle equation.

Now we must find the length of the diameter so that we can know what our radius is. The distance between any two points is:

d^2=(x2-x1)^2+(y2-y1)^2 so

d^2=(-1-3)^2+(-4-2)^2

d^2=4^2+6^2

d^2=16+36

d^2=52

Note that this distance squared is our diameter squared.

Since d=2r

d^2=4r^2

r^2=d^2/4, we showed that d^2=52 so

r^2=52/4

r^2=13 (r=√13, but we need r^2 for our equation anyway)

So now we can fill out our circle equation because we know the center is at (1,-1) are r^2=13

(x-1)^2+(y+1)^2=13