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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 $C$ ?
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Verified Answer
The correct answer is:
$(2,4)$
Equation of circle
$\mathrm{x}^2+(\mathrm{y}-(6-\mathrm{r}))^2=\mathrm{r}^2$
touches $\sqrt{3} \mathrm{x}-\mathrm{y}=0$
$\begin{aligned}
& \mathrm{p}=\mathrm{r} \\
& \frac{|0-(6-\mathrm{r})|}{2}=\mathrm{r} \\
& |\mathrm{r}-6|=2 \mathrm{r} \\
& \mathrm{r}=2
\end{aligned}$
$\therefore$ Circle $\mathrm{x}^2+(\mathrm{y}-4)^2=4$
$(2,4)$ Satisfies this equation
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