determine the differential equation giving the slope of the tangent line at point (x,y) for the given family of curves. x^2+y^2=2cx

\[y'=\frac{(c-x) }{y} \]

the answer is y'=(y\[y'=(y^2-x^2)/2xy\] I just don't know how they got this. I do not know where to even begin.

From Mathematica:\[\text{DSolve}\left[y'[x]= \frac{\left(y[x]^2-x^2\right)}{2x y[x]},y[x],x\right]\]\[\left\{\left\{y[x]\to -\sqrt{-x^2+x C[1]}\right\},\left\{y[x]\to \sqrt{-x^2+x C[1]}\right\}\right\} \]Take the second solution and square each side:\[y[x]^2=-x^2+x C[1]\]Move the -x^2 from the RHS to the LHS.\[y[x]^2 + x^2 = x C[1]\]If the same procedure is applied to my answer then the following is the result:\[y[x]^2+x^2\text{= }2 c x+2 C[1]\]where C[1] is zero. I am not sure that their expression is totally correct.

y'=(c-y)/y came from taking the Total Derivative of the problem expression and solving for dy/dx after setting dc to zero.

the 2 answers are equivalent....the textbook answer just replaced c with (x^2+y^2)/2x which comes from solving original equation for c. \[\frac{c-x}{y} = \frac{\frac{x^{2}+y^{2}}{2x} -x}{y} = \frac{\frac{x^{2}+y^{2}-2x^{2}}{2x}}{y} = \frac{y^{2}-x^{2}}{2xy}\]

@ dumbcow Thank you for the explanation.

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