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 $C$ ?

Question

Aicharya · Accepted Answer

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

Let $C$ be the circle of minimum area touching the parabola $y=6-x^2$ and the lines $y=\sqrt{3}|x|$. Then,…

Solution