1. Use symmetry to locate the center of the circle

The parabola $y=6-x^2$ and the pair of lines $y=\sqrt{3}|x|$ are symmetric about the $y$-axis. Hence the minimum-area circle touching all these curves must also be symmetric about the $y$-axis.

So let the center of the circle be $(0,k)$ and radius be $r$.

2. Condition for touching the lines

The two lines are $y=\sqrt{3}x \quad (x\ge 0), \qquad y=-\sqrt{3}x \quad (x\le 0).$ Equivalently, $\sqrt{3}x-y=0, \qquad -\sqrt{3}x-y=0.$

The perpendicular distance from $(0,k)$ to the line $\sqrt{3}x-y=0$ is $\frac{|\sqrt{3}(0)-k|}{\sqrt{3+1}}=\frac{|k|}{2}.$ Since the circle touches both lines, this distance equals the radius: $r=\frac{|k|}{2}.$

Because the relevant circle is above the origin, take $k>0$, so $r=\frac{k}{2}.$

3. Condition for touching the parabola

We need the circle to be tangent to the parabola. The circle is $x^2+(y-k)^2=r^2.$ A point on the parabola is $(x,6-x^2).$ So the squared distance from the center $(0,k)$ to a point on the parabola is $D^2=x^2+(6-x^2-k)^2.$ For tangency, the minimum value of this expression must equal $r^2$.

Let $f(x)=x^2+(6-k-x^2)^2.$ Put $a=6-k.$ Then $f(x)=x^2+(a-x^2)^2=x^4+(1-2a)x^2+a^2.$

Differentiate: $f'(x)=4x^3+2(1-2a)x=2x\bigl(2x^2+1-2a\bigr).$ Hence critical points are given by $x=0$ or $2x^2+1-2a=0 \implies x^2=a-\frac12.$

Now,

• at $x=0$, we get $f(0)=a^2$,
• at $x^2=a-\frac12$, we get $a-x^2=a-\left(a-\frac12\right)=\frac12,$ so $f_{\min}=x^2+\left(\frac12\right)^2=a-\frac12+\frac14=a-\frac14.$

Thus when $a\ge \tfrac12$, the minimum distance squared is $r^2=a-\frac14=6-k-\frac14=\frac{23}{4}-k.$

But from tangency to the lines, $r=\frac{k}{2} \implies r^2=\frac{k^2}{4}.$ Therefore, $\frac{k^2}{4}=\frac{23}{4}-k.$ Multiply by $4$: $k^2=23-4k,$ $k^2+4k-23=0.$ So $k=-2\pm 3\sqrt{3}.$ Since $k>0$, $k=-2+3\sqrt{3}.$ Hence $r=\frac{k}{2}=\frac{-2+3\sqrt{3}}{2}.$

4. Find the tangent point on the parabola

Recall $x^2=a-\frac12=(6-k)-\frac12=\frac{11}{2}-k.$ Substitute $k=-2+3\sqrt3$: $x^2=\frac{11}{2}-(-2+3\sqrt3)=\frac{15}{2}-3\sqrt3.$ This is not directly among the options, so instead let us use the normal condition more cleanly.

For parabola $y=6-x^2$, $\frac{dy}{dx}=-2x.$ So slope of normal at $(x,6-x^2)$ is $m_n=\frac{1}{2x}$ (for $x\neq 0$).

Since the radius to the tangent point is normal to the parabola, the center $(0,k)$ lies on this normal: $\frac{k-(6-x^2)}{0-x}=\frac{1}{2x}.$ So $\frac{k-6+x^2}{-x}=\frac{1}{2x}.$ Multiply by $x$: $-(k-6+x^2)=\frac12,$ $k=\frac{11}{2}+x^2.$ This sign is inconsistent because center should lie below the tangent point for positive $x$; let us compute carefully.

Slope of tangent is $-2x$, so slope of normal is $m_n=-\frac{1}{-2x}=\frac{1}{2x}.$ The slope from $(x,6-x^2)$ to center $(0,k)$ is $\frac{k-(6-x^2)}{0-x}=\frac{k-6+x^2}{-x}.$ Set equal: $\frac{k-6+x^2}{-x}=\frac{1}{2x}.$ Thus $-(k-6+x^2)=\frac12,$ $k=\frac{11}{2}+x^2.$ But this would make $k>\frac{11}{2}$, impossible since the circle must lie below vertex $(0,6)$. Therefore the correct normal slope should be $m_n=-\frac{1}{-2x}=\frac{1}{2x},$ but for the left branch one may get sign changes. Instead use the derivative minimization relation:

From $2x^2+1-2a=0,$ with $a=6-k$, we have $2x^2+1-2(6-k)=0,$ $2x^2+1-12+2k=0,$ $2x^2=11-2k.$ So $x^2=\frac{11-2k}{2}.$ Substitute $k=-2+3\sqrt3$: $x^2=\frac{11-2(-2+3\sqrt3)}{2}=\frac{15-6\sqrt3}{2}.$ Then $y=6-x^2=6-\frac{15-6\sqrt3}{2}=\frac{-3+6\sqrt3}{2}.$

This is the tangency point, but we still need which given option lies on the circle.

5. Test the given points in the circle equation

Circle equation: $x^2+(y-k)^2=r^2, \qquad r=\frac{k}{2}, \quad k=-2+3\sqrt3.$ So $x^2+\left(y+2-3\sqrt3\right)^2=\left(\frac{-2+3\sqrt3}{2}\right)^2.$

Instead of expanding irrationally for every option, use the equivalent form from $x^2+y^2-2ky+k^2=r^2=\frac{k^2}{4},$ so $x^2+y^2-2ky+\frac{3k^2}{4}=0.$ Now from $k^2+4k-23=0 \implies \frac{3k^2}{4}=\frac{3(23-4k)}{4}=\frac{69}{4}-3k.$ Thus $x^2+y^2-2ky+\frac{69}{4}-3k=0.$

Now test options numerically using $k=-2+3\sqrt3\approx 3.196$ and $r\approx 1.598.$

• A $(1,2)$: $\text{distance from center}^2=1^2+(2-3.196)^2\approx 1+1.430=2.430,$ while $r^2\approx 2.554.$ Not equal.

• B $(2,2)$: $4+(2-3.196)^2\approx 4+1.430=5.430,$ not equal.

• C $(1,1)$: $1+(1-3.196)^2\approx 1+4.822=5.822,$ not equal.

• D $(2,4)$: $4+(4-3.196)^2\approx 4+0.646=4.646,$ not equal.

So something is wrong in assuming the minimum circle is tangent to the parabola at a non-vertex point.

6. Check tangency at the vertex of the parabola

At the vertex $(0,6)$, the tangent is horizontal, so a circle centered on the $y$-axis can touch there naturally.

If the circle touches the parabola at $(0,6)$, then the radius is $r=6-k.$ But from touching the lines, we also have $r=\frac{k}{2}.$ Therefore, $6-k=\frac{k}{2},$ $12-2k=k,$ $3k=12,$