Let C be the circle of minimum area touching the parabola y=6-x^2 and the lines [y=√3|x|].

Question

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 $C$ ?

Accepted Answer

Let centre be (0, k)

Now radius is $r=6-k$

Also, $$6-k=\left|\frac{k}{2}\right|$$

$$\begin{aligned}
& \Rightarrow 6-k=\frac{k}{2} \
& \Rightarrow 12-2 k=k \
& \Rightarrow k=4
\end{aligned}$$

Radius, $r=6-4=2$

So circle will be

$$\begin{aligned}
& (x)^2+(y-k)^2=4 \
& x^2+(y-4)^2=4
\end{aligned}$$

$(2,4)$ satisfies this equation.

Solutions · 1