The radius of the circle C is ___ .

The circles in the family lie completely inside the region
$$R=\Bigl\{(x,y):x\ge 0\ \text{and}\ y^{2}\le 4-x\Bigr\}.$$
The boundary of the region is the left-opening parabola $$y^{2}=4-x$$ together with the y-axis $$x=0$$.

Because every required circle must have its centre on the x-axis, write the centre as $$\,(h,0)\,(h\ge 0)$$ and its radius as $$r$$. The circle is therefore
$$\bigl(x-h\bigr)^{2}+y^{2}=r^{2}.$$ To be completely contained in $$R$$ every point $$(x,y)$$ on the circle must satisfy both constraints of $$R$$. At the extreme points of the circle these give three elementary (necessary) inequalities:

• Leftmost point $$(h-r,0)$$ must obey $$x\ge 0\;$$ $$\Rightarrow$$ $$h-r\ge 0\;,$$ hence $$r\le h.$$

• Rightmost point $$(h+r,0)$$ must satisfy $$y^{2}\le 4-x$$ with $$y=0\;$$ $$\Rightarrow$$ $$h+r\le 4.$$

• Top (and bottom) point $$(h,r)$$ must satisfy $$r^{2}\le 4-h.$$

Although these three conditions are necessary, they are not sufficient: the whole curved boundary of the circle must still lie inside the parabola. At the largest possible radius the circle will be tangent to the parabola $$y^{2}=4-x$$ at some point, say $$(\alpha,\beta)$$ with $$\beta\ne 0$$ (otherwise the circle would shrink).

Since $$\alpha$$ and $$\beta$$ lie on the parabola, $$\alpha=4-\beta^{2}\qquad -(1).$$

The slope of the tangent to the parabola is obtained by implicit differentiation of $$y^{2}=4-x$$:
$$2y\,\frac{dy}{dx}=-1\ \Longrightarrow\ \frac{dy}{dx}=-\frac{1}{2y}.$$ Hence the slope of the normal at $$(\alpha,\beta)$$ is $$m_{n}=2\beta.$$

Because the centre $$(h,0)$$ of the circle lies on this normal, the slope of the line joining $$(h,0)$$ and $$(\alpha,\beta)$$ equals $$2\beta$$:
$$\frac{\beta-0}{\alpha-h}=2\beta\quad\Longrightarrow\quad \alpha-h=\frac12.$$ Therefore $$h=\alpha-\frac12=4-\beta^{2}-\frac12=\frac72-\beta^{2}\qquad -(2).$$

The radius of the circle is the distance between its centre and the tangency point: $$r=\sqrt{(\alpha-h)^{2}+\beta^{2}}=\sqrt{\left(\frac12\right)^{2}+\beta^{2}}=\sqrt{\beta^{2}+\frac14}\qquad -(3).$$

For the circle to stay on the right of the y-axis we must also have the leftmost point exactly on (or to the right of) the axis when the radius is maximised. Imposing the boundary condition $$h-r=0$$ (touching the y-axis) and using (2) and (3):
$$\frac72-\beta^{2}-\sqrt{\beta^{2}+\frac14}=0.$$ Let $$z=\beta^{2}\;(z\ge 0)$$. Squaring once removes the root:

$$\Bigl(\tfrac72-z\Bigr)^{2}=z+\tfrac14$$ $$\Longrightarrow\ z^{2}-8z+12=0$$ $$\Longrightarrow\ z=6\quad\text{or}\quad z=2.$$

The choice $$z=6$$ would give $$\alpha=4-6=-2$$ and $$h=\tfrac72-6=-\tfrac52\lt 0$$, placing both the centre and the tangency point outside $$R$$, so it is rejected. Hence $$\beta^{2}=2\;\Longrightarrow\;\beta=\pm\sqrt2,\qquad \alpha=4-2=2,\qquad h=\frac72-2=\frac32.$$

Finally, substitute $$\beta^{2}=2$$ into (3):
$$r=\sqrt{2+\frac14}=\sqrt{\frac94}=\frac32=1.5.$$

All earlier necessary conditions are satisfied: leftmost point $$h-r=0$$ lies on the y-axis, rightmost point $$h+r=3\lt 4,$$ and top point $$(\tfrac32,\tfrac32)$$ obeys $$y^{2}=2.25\le 2.5.$$ Because the circle is tangent to the parabola and to the y-axis, any larger radius would push part of the circle outside the allowable region. Therefore this circle indeed has the largest possible radius.

Largest radius of the circle $$C$$: $$\displaystyle r_{\max}=1.5.$$

Hence, rounded to two decimal places, the required radius lies in the interval 1.49 - 1.51.