Asked by Marie P.
Find the equation of the circle with endpoints of diameter (-1,3) and (7, -5)
Answers
Answered by
bobpursley
center must be x,y, or (3,1)
diameter is more compliated:
diameter=sqrt[(-1-7)^2 + (3+5)^2 ]
radius is half that
(x-3)^2+(y-1)^2=radius^2 will work.
check that math, it is easy to err when typing
diameter is more compliated:
diameter=sqrt[(-1-7)^2 + (3+5)^2 ]
radius is half that
(x-3)^2+(y-1)^2=radius^2 will work.
check that math, it is easy to err when typing
Answered by
Reiny
or
Once you have bobpursley's equation of
(x-3)^2 + (y+1)^2 = r^2
sub in one of the points on the circle, I will use (-1,3)
(-1-3)^2 + (3+1)^2 = r^2
16 + 16 = r^2 = 32
(x-3)^2 + (y-1)^2 = 32
I noticed that bob had the centre as (3,1),
should have been (3,-1)
Once you have bobpursley's equation of
(x-3)^2 + (y+1)^2 = r^2
sub in one of the points on the circle, I will use (-1,3)
(-1-3)^2 + (3+1)^2 = r^2
16 + 16 = r^2 = 32
(x-3)^2 + (y-1)^2 = 32
I noticed that bob had the centre as (3,1),
should have been (3,-1)
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