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 $C$?

• A. $(2, 4)$
• B. $(1, 2)$
• C. $(2, 2)$
• D. $(1, 1)$

Answer 1

Answer: (A)

$(2, 4)$

Answer 2

The equation of the circle is:

$x^2 + (y - (6 - r))^2 = r^2,$

where the center is $(0, 6 - r)$ and radius is $r$.

Step 1: Condition for tangency with $y = \sqrt{3}|x|$: The perpendicular distance from the center $(0, 6 - r)$ to the line $y = \sqrt{3}|x|$ must equal the radius $r$.
For the line $y = \sqrt{3}x$, the distance is: $\frac{|0 - (6 - r)|}{\sqrt{1^2 + (\sqrt{3})^2}} = r.$ Simplify: $\frac{|6 - r|}{2} = r.$

Step 2: Solve for $r$:
Case 1: $6 - r = 2r \implies 6 = 3r \implies r = 2.$
Case 2: $6 - r = -2r \implies 6 = -r \implies r = -6$ (not valid as $r > 0$).
Hence, $r = 2$.

Step 3: Equation of the circle: Substituting $r = 2$, the center becomes $(0, 6 - 2) = (0, 4)$.
The equation of the circle is: $x^2 + (y - 4)^2 = 4.$

Step 4: Check which point lies on the circle: Substituting $(2, 4)$ into the equation:
$2^2 + (4 - 4)^2 = 4 \implies 4 + 0 = 4.$ Thus, $(2, 4)$ lies on the circle.