Let $C$ be the circle of minimum area touching the parabola $y=6-x^2$ and the lines $y=\sqrt{3}|x|$. Then which one of the following points lies on the circle $C$?
Step-by-Step Solution
Key Concept: By symmetry, circle centre is on $y$-axis at $(0,6-r)$. It touches $y=\sqrt{3}x$ (i.e. $\sqrt{3}x-y=0$): distance $=\frac{|0-(6-r)|}{2}=r\Rightarrow|r-6|=2r$.
$r=2$, centre $(0,4)$. Circle $x^2+(y-4)^2=4$. Point $(2,4)$: $4+0=4$ ✓.
Correct Answer: 4