Asked by Lleyton
Find the equation of a circle passing through the origin and the points where the line 3x + 4y=12 meets the axes of the co-ordinates..
Answers
Answered by
Steve
the circle passes through
(0,0), (4,0), (0,3)
If the center is (h,k) and the radius is r, then
h^2 + k^2 = r^2
(4-h)^2 + k^2 = r^2
h^2 + (3-k)^2 = r^2
The circle is thus
(x-2)^2 + (y-3/2)^2 = (5/2)^2
see
http://www.wolframalpha.com/input/?i=%28x-2%29^2+%2B+%28y-3%2F2%29^2+%3D+%285%2F2%29^2
(0,0), (4,0), (0,3)
If the center is (h,k) and the radius is r, then
h^2 + k^2 = r^2
(4-h)^2 + k^2 = r^2
h^2 + (3-k)^2 = r^2
The circle is thus
(x-2)^2 + (y-3/2)^2 = (5/2)^2
see
http://www.wolframalpha.com/input/?i=%28x-2%29^2+%2B+%28y-3%2F2%29^2+%3D+%285%2F2%29^2
Answered by
Lleyton
How can we say that the circle passes through (4,0),(0,3)???
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