JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 4)
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 $$C$$ ?
$$(1,2)$$
$$(2,2)$$
$$(1,1)$$
$$(2,4)$$
Explanation
Let centre be (0, k)
_6th_April_Morning_Shift_en_4_1.png)
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.
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