Let equation of the circle be

x² + (y – t)² = r²              ......(1)

y = 4 – x²                       ......(2)

 $\text{t }=\mathrm{r}\sqrt{2}$                                   .....(3)

From (1) and (2)

$\text{4 }–\text{ y }+ {(\mathrm{y}-\mathrm{r}\sqrt{2})}^2={\text{ r}}^{\text{2}}$

${\text{y}}^{\text{2}} – \mathrm{y}(2\sqrt{2}\mathrm{r}+1)+{\text{ r}}^{\text{2}} +\text{ 4 }=0$

Now,     D = 0                           ⇒        ${(2\sqrt{2}\mathrm{r}+1)}^2–4({\mathrm{r}}^2+4)=0$       

4r² +  $4\sqrt{2}\mathrm{r}$ – 15 = 0                  ⇒         4r² + $2·2\sqrt{2}\mathrm{r}$ + 2 = 17

$\Rightarrow {(2\mathrm{r}+\sqrt{2})}^2=\text{ 17}$                           $\Rightarrow 2\mathrm{r}+\sqrt{2}=\sqrt{17}$

$\text{r }= \frac{\sqrt{17}-\sqrt{2}}{2}\cong 1.3$

Again, when the circle touches the parabola at (0, 4) then

4 – t = r             ...(i)

$\mathrm{t}=\mathrm{r}\sqrt{2}$           ...(ii)

from (i) and (ii)

$\text{4 }–\text{ r}\sqrt{2}=\text{ r}\Rightarrow \text{r }=\frac{4}{\sqrt{2}+1}$

$\Rightarrow \text{r }=\text{ 4}(\sqrt{2}–1).\simeq 1.656.$