Form differential equation of circle having centre on y axis and radius is 3 units I got the answer but its not matching. I got $x+y\frac{dy}{dx}-3\frac{dy}{dx} =0$. Please suggest is that correct.
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First of all, this does not look like an equation. – Shailesh Sep 25 '15 at 05:42
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$$x^{2}+(y-b)^{2}=9$$ By differentiating. $$2x+2(y-b)(y')=0$$ Now make value of (y-b) and put it in equation of circle $$(y-b) = \frac{-x}{y'}$$ And you will get $$(x^{2}-9)(y'^{2})+x^{2}=0$$
piyush
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Let $(0,c)$ be the center of the circle and $r = 3, c > 0 \Rightarrow x^2+(y-c)^2 = 3^2=9\Rightarrow 2x+2(y-c)\dfrac{dy}{dx}=0\Rightarrow x + (y-c)\dfrac{dy}{dx}=0$
DeepSea
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Just a doubt. Wouldn't you have to differentiate it once again and eliminate c in order to get a D.E. without c ? – Shailesh Sep 25 '15 at 05:46
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You shouldn't eliminate $c$ since it is a constant. If you want to eliminate then you eliminate a variable $x$ or $y$. – DeepSea Sep 25 '15 at 05:48
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your answer is just as i have done as the centre is on y axis and raius is 3 c=3 . and thats not matching. – Adesh Tamrakar Sep 25 '15 at 05:50
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c is a parameter here ... not really a constant... more like a variable constant. Anyways, I might post it and see what others think. Thanks – Shailesh Sep 25 '15 at 05:51
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I think shailesh is right as its not given that the circle passes through origin. – Adesh Tamrakar Sep 25 '15 at 05:54
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A1DHTH is correct. You are confusing $c$ with the value of the radius. $c$ is not the radius, it is an arbitrary coordinate for the center's y-value. – zahbaz Sep 25 '15 at 05:57
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Slope of normal line between $(0,3) - (x,y) =\dfrac{y-3}{x} $
So, slope of tangent line $\dfrac {dy}{dx} @( x,y ) $ is its negative reciprocal $ \dfrac {dy}{dx}=\dfrac{-x}{y-3} $
Simplifying, it correctly comes to what you have given.
Narasimham
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If you want to use a parametric equation, $$\dot{x} = y$$ $$\dot{y}=-x$$ with $x(0) = x_0$ and $y(0) = y_0$ is also a circle centered at the origin with radius $\sqrt{x_0^2+y_0^2}$. You can change the center by simply translating $x$ and $y$.
Calculon
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