Explanation:
(A) : Let, (h, k) be the centre of the circle. Then, equation of circle having radius ' \(a\) ' be
\((x-h)^{2}+(y-k)^{2}=a^{2} \tag{i}\)
Differentiating both the sides, we get-
\(2(x-h)+2(y-k) \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=-\left(\frac{x-h}{y-k}\right) \tag{ii}\)
Again differentiating both the sides, we get-
\((y-k) \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=-1 \Rightarrow(y-k)=\frac{-1-\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\)
From equation (i), we have
\(\frac{(\mathrm{x}-\mathrm{h})^{2}}{(\mathrm{y}-\mathrm{k})^{2}}+1=\frac{\mathrm{a}^{2}}{(\mathrm{y}-\mathrm{k})^{2}} \Rightarrow(\mathrm{y}-\mathrm{k})^{2}\left[\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+1\right]=\mathrm{a}^{2}\)
\(a^{2}=\left[\frac{-1-\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\right]^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]\)
\(a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=\left(\left(\frac{d y}{d x}\right)^{2}+1\right)^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]\)
Hence, order is 2.
