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Question 1124951: please help me solve this
find the equation for the circle with a diameter whose endpoints are
(2,-2) and (1,5)
i used this equation (x-h)^2+(y-k)^2=r^2
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! please help me solve this
find the equation for the circle with a diameter whose endpoints are
(2,-2) and (1,5)
i used this equation (x-h)^2+(y-k)^2=r^2
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How did you use the equation?
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The center of the circle is the midpoint of the 2 points, (h,k)
The radius is the distance from the midpoint to either point.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! your diameter goes between the points (2,-2) and (1,5)
the distance between those 2 points is sqrt((x2-x1)^2 + (y2-y1)^2)
let (2,-2) = (x1,y1) and let (1,5) = (x2,y2).
the distance between those two points becomes equal to sqrt((1-2)^2 + (5--2)^2)
that becomes sqrt(-1^2 + 7^2) = sqrt(50)
that's the length of the diameter.
the length of the radius is half that, so the length of the radius is sqrt(50) / 2.
the center of the circle is the midpoint of the diameter.
the formula for that is (x1 + x2) / 2, (y1 + y2) / 2)
since (2,-2) = (x1,y1) and (1,5) = (x2,y2), the midpoint formula becomes:
((2+1)/2,(-2+5)/2) = (3/2,3/2)
that's the center of the circle, therefore (h,k) = (3/2,3/2).
r = sqrt(50)/2 which makes r^2 = 50/4 = 12.5
the formuls of the circle is (x-h)^2+(y-k)^2=r^2
that becomes (x - 3/2)^2 + (y - 3/2)^2 = 12.5
here's the graph of that equation.
the blue line is the line that contains the diameter of the circle.
the circle itself is also in blue.
the orange lines are only there to allow the intersections to be shown.
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