find the differential equation of the circle in xy plane
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Equation of a circle is given by -
(x - a)^2 + (y - b)^2 = R^2.
Now, we know that ,
d/dx (x - a)^2 = 2(x - a).
d/dy (y - b)^2 = 2(y - b).
Therefore, the differential equation of circle is :
2 (x - a) dx + 2 (y - b) dy = 0.
(x-a)dx + (y-b)dy = 0.
where, (a, b) is the center of the circle!
Thanks!
(x - a)^2 + (y - b)^2 = R^2.
Now, we know that ,
d/dx (x - a)^2 = 2(x - a).
d/dy (y - b)^2 = 2(y - b).
Therefore, the differential equation of circle is :
2 (x - a) dx + 2 (y - b) dy = 0.
(x-a)dx + (y-b)dy = 0.
where, (a, b) is the center of the circle!
Thanks!
Sbharyanvi:
Its not complete solution.please eliminate a and b.
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