Question 816642: Please can you help solve this problem for me. The point (x,y) lies on a cirle which has the segment from (0,-1) to (1/3, 4/3) as a diameter. Show this is algebraic equation
Answer by LinnW(1048)   (Show Source):
You can put this solution on YOUR website! Here we are looking for a circle with its center at the midpoint between
(0,-1) and (1/3,4/3). The midpoint is ( (0+1/3)/2 , (-1+4/3)/2 ).
( 1/6 , 1/6 ). We also need the length of the radius of the circle
which has as diameter (0,-1) and (1/3,4/3). The length of the diameter
is sqrt( (y2-y1)^2 + (x2-x1)^2 ).
So we want sqrt ( ( 1/3 - 0 )^2 + ( 4/3 - (-1) )^2 )
= sqrt ( ( 1/3 )^2 + ( 4/3 + 1 )^2 )
= sqrt ( 1/9 + (7/3)^2 )
= sqrt ( 1/9 + 49/9 )
= sqrt ( 50/9 )
The radius is half of this or 25/9 .
The standard form for a circle is (x-h)^2 + (y-k)^2 = r^2
For our application h= 1/6 and k= 1/6 , with r = 25/9 r^2 = 625/81
So (x- 1/6)^2 + (y- 1/6)^2 = 625/81
Double check this. Plotting the above equation
http://www.wolframalpha.com/input/?i=%28x-+1%2F6%29^2+%2B+%28y-+1%2F6%29^2+%3D+625%2F81
|
|
|
