Find max. and min. value of 'r' 
Find $\max$ and $\min$ values of $r$ if
  $$r^2 - 4x + 3 = 0$$
  where $r= \sqrt{x^2+y^2}$  

Tried differentiating but didn't get any result for max. value of $ \sqrt{x^2+y^2}$

If $r^2-4x+3=0$ then $y^2=4x-3-x^2=-(x-3)(x-1)$.
Since $y^2\ge0$, you must have $x \in [1,3]$.
Then, $r=\sqrt{4x-3}$, so $r$ is minimum when $\sqrt{4x-3}$ is minimum, that is $x=1$ (notice $x \to \sqrt{4x-3}$ is an incresaing function) and $y=0$, and $r=1$.
And $r$ is maximum when $\sqrt{4x-3}$ is maximum, thus $x=3$, $r=\sqrt{4\cdot3-3}=3$ and $y=0$.
