The shortest distance between the ellipse ${\mathrm{x}}^2+2{\mathrm{y}}^2=2$and the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-3\mathrm{x}-2\sqrt{2}\mathrm{y}+4=0$  is

The normal at $\mathrm{\theta }$ to ellipse is $\sqrt{2}\mathrm{xsec}\mathrm{\theta }-\mathrm{ycosec}\mathrm{\theta }=1$ 

 It is passes through centre of circle    
$\left(\frac{3}{2},\sqrt{2}\right)\Rightarrow \frac{3}{\sqrt{2}}\sec \mathrm{\theta }-\sqrt{2}\mathrm{cosec}\mathrm{\theta }=1\Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{4}$
Then point on ellipse is  $\left(1,\frac{1}{\sqrt{2}}\right)$
  $\therefore$shortest distance =CP– radius  $=\frac{\sqrt{3}}{2}-\frac{1}{2}$