SOLUTION: If the endpoints of a diameter of a circle are (4,3)and (-3,5), what is the equation of the circle? This is my answer is this right, if not what is? Answer (5-3)^2+(-3-4)^2 =2

Algebra ->  Trigonometry-basics -> SOLUTION: If the endpoints of a diameter of a circle are (4,3)and (-3,5), what is the equation of the circle? This is my answer is this right, if not what is? Answer (5-3)^2+(-3-4)^2 =2      Log On


   

Question 460962: If the endpoints of a diameter of a circle are (4,3)and (-3,5), what is the equation of the circle?
This is my answer is this right, if not what is?
Answer
(5-3)^2+(-3-4)^2
=2^2+(-7)^2
=4+49
=53
radius=53/2
midpoint : (1/2, 4)
equation of the circle=(x-1/2)^2+(y-4)^2=53/4
Found 2 solutions by jim_thompson5910, Gogonati:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The radius is actually r=sqrt%2853%29%2F2, which will make r%5E2=53%2F4. You forgot to take the square root of 53 in the last step.



Anyways, the equation you have is the correct one nonetheless. Good job.

Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
Since we know the coordinates of the endpoints of the diameter we can find the coordinate of the center:
h=%28-3%2B4%29%2F2=1%2F2 and k=%285%2B3%29%2F2=4, thus the circle is centered at
(1/2, 4). Now we need to find out the radius of the circle:
r%5E2=%28-3-4%29%5E2%2B%285-3%29%5E2=49%2B4=53. As we know the equation of the circle in
standard form is : %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2, substituting our values we get:
%28x-1%2F2%29%5E2%2B%28y-4%29%5E2=53, which is the equation of our circle.