Find an equation of the circle whose endpoints of a diameter are (−3,6) and (5,−14).
I tried to figure it out, but I'm not sure if my answer is correct.
I got..
(x-2)^2 + (y + 4)^2 = 116
Any help would be great. Thank you!
the centre must be the midpoint of the diameter which would give you (1, -4)
so the equation must be
(x-1)^2 + (y+4)^2 = r^2
sub in one of the points on the circle , I will use (-3,6)
(-4)^2 + (10)^2 = r^2
r^2 = 16+100= 116
so (x-1)^2 + (y+4)^2 = 116
check by subbing in the other endpoint, it works
You made one small error
Thank you very much your help. I appreciate it! :)
To find the equation of the circle whose endpoints of a diameter are given, you can use the midpoint formula and the distance formula.
Step 1: Find the coordinates of the midpoint of the diameter.
The midpoint formula is given by:
x-coordinate of midpoint = (x1 + x2)/2
y-coordinate of midpoint = (y1 + y2)/2
Using the given endpoints: (-3, 6) and (5, -14)
x-coordinate of midpoint = (-3 + 5)/2 = 2/2 = 1
y-coordinate of midpoint = (6 + (-14))/2 = -8/2 = -4
So, the midpoint of the diameter is (1, -4).
Step 2: Find the length of the radius.
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Using the given endpoints:
d = sqrt((5 - (-3))^2 + (-14 - 6)^2)
= sqrt((5 + 3)^2 + (-20)^2)
= sqrt(8^2 + (-20)^2)
= sqrt(64 + 400)
= sqrt(464)
= 4sqrt(29)
So, the radius of the circle is 4sqrt(29).
Step 3: Write the equation of the circle.
The equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
Using the midpoint (1, -4) and the radius 4sqrt(29):
(x - 1)^2 + (y + 4)^2 = (4sqrt(29))^2
(x - 1)^2 + (y + 4)^2 = 16 * 29
(x - 1)^2 + (y + 4)^2 = 116
Therefore, the correct equation of the circle with the given endpoints as the diameter is:
(x - 1)^2 + (y + 4)^2 = 116.
Your answer, (x - 2)^2 + (y + 4)^2 = 116, is very close. It seems that you made a small mistake in the x-coordinate of the midpoint. Instead of (x - 2)^2, it should be (x - 1)^2.