Discussion
The differential equation of family of circles of radius \(r\) whose center lie on \(x\)-axis is :
Answer: D
The differential equation is \(y^{2}\left[\left(\frac{dy}{dx}\right)^{2} + 1\right] = r^{2}\).
A circle of radius \(r\) with centre \((h, 0)\) on the \(x\)-axis is \((x - h)^{2} + y^{2} = r^{2}\), with one arbitrary constant \(h\). Differentiate to eliminate it:
\[ \begin{aligned} 2(x - h) + 2y\,y' &= 0 \\ x - h &= -y\,y' \\ y^{2} y'^{2} + y^{2} &= r^{2} \\ y^{2}\left(y'^{2} + 1\right) &= r^{2} \end{aligned} \]
Option A, \(y y' + y^{2} = r^{2}\), forgets to square \(y y'\). Option B, \(y[(y')^{2} + 1] = r^{2}\), has only one power of \(y\) where the substitution produces \(y^{2}\), and option C, \(y^{2}[y' + 1]\), forgets to square the derivative.
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