HCU MCA 2013 Paper

(a) Given equation of circle,
${\left(\mathit{x}-\mathit{h}\right)}^2+{\left(\mathit{y}-\mathit{k}\right)}^2=1...\left(\mathrm{i}\right)$
On differentiating w.r.t x, we get,
$\Rightarrow \mathrm{}2\left(\mathit{x}-\mathit{h}\right)+2\left(\mathit{y}-\mathit{k}\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0$
$\Rightarrow \left(\mathit{x}-\mathit{h}\right)+\left(\mathit{y}-\mathit{k}\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0...\left(\mathrm{ii}\right)$
Again, diffierentiating w.r.t x, we get
$\Rightarrow 1+\left(\mathit{y}-\mathit{k}\right)\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}+\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}.\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0$
$\Rightarrow \frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}\left(\mathit{y}-\mathit{k}\right)+{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2=-1...\left(\mathrm{iii}\right)$
From Eqs. (ii) and (iii), we obtain
$\left(\mathit{y}-\mathit{K}\right)=-\frac{1+{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2}{\mathrm{}\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}}$
and $\left(\mathit{x}-\mathit{h}\right)=\mathrm{}\frac{\left\{1+{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2\right\}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}}{\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}}$
On substituting these values in the Eq. (i), we get
${\left[1+{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2\right]}^3={\left(\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}\right)}^2$
which is the required differential equation.