Dear experts is the answer right . the question is form a differential equation of family of circles passing through the points(a,0) and (-a,0)

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$$\begin{gathered} \mathrm{As} \mathrm{there} \mathrm{is} \mathrm{only} \mathrm{one} \mathrm{variable} \mathrm{k} , \mathrm{hence} \mathrm{we} \mathrm{must} \mathrm{differentiate} \mathrm{only} \mathrm{once}. \\ (\mathrm{a},0)\mathrm{and}(-\mathrm{a},0) \\ \mathrm{midpoint}\mathrm{is}(0,0) \\ \mathrm{The}\mathrm{center}\mathrm{of}\mathrm{these}\mathrm{circles}\mathrm{necessarily}\mathrm{lie}\mathrm{on}\mathrm{the}\mathrm{y}\mathrm{axis},\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{chord}\mathrm{passes}\mathrm{through}\mathrm{centre} \\ \mathrm{This}\mathrm{implies}\mathrm{that}\mathrm{their}\mathrm{center}\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{as}(\mathrm{h},\mathrm{k})\mathrm{where}\mathrm{h}=0 \\ \mathrm{By}\mathrm{Pythagoras}\mathrm{theorem},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{such}\mathrm{circle}\mathrm{can}\mathrm{be}\mathrm{given}\mathrm{by}\mathrm{r}=\sqrt{{\mathrm{a}}^2+{\mathrm{k}}^2} \\ \mathrm{Now}\mathrm{just}\mathrm{substitute}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{standard}\mathrm{equation}: \\ {\mathrm{x}}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{a}}^2+{\mathrm{k}}^2 \\ \mathrm{differentiating}\mathrm{we}\mathrm{get} \\ 2\mathrm{x}+2(\mathrm{y}-\mathrm{k})\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}-\mathrm{k}} \end{gathered}$$

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