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

The Correct Option is A

The circle's equation is:

\[ x^2 + (y - (6 - r))^2 = r^2, \]

with center \((0, 6 - r)\) and radius \(r\).

Step 1: Tangency condition with \(y = \sqrt{3}|x|\): The 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 formula yields: \[ \frac{|0 - (6 - r)|}{\sqrt{1^2 + (\sqrt{3})^2}} = r. \] Simplifying gives: \[ \frac{|6 - r|}{2} = r. \]

Step 2: Solving for \(r\):
Case 1: \(6 - r = 2r \implies 6 = 3r \implies r = 2.\)
Case 2: \(6 - r = -2r \implies 6 = -r \implies r = -6\) (rejected as \(r>0\)).
Therefore, \(r = 2\).

Step 3: Circle equation: With \(r = 2\), the center is \((0, 6 - 2) = (0, 4)\).
The circle's equation is: \[ x^2 + (y - 4)^2 = 4. \]

Step 4: Point verification: Substituting \((2, 4)\) into the equation:
\[ 2^2 + (4 - 4)^2 = 4 \implies 4 + 0 = 4. \] This confirms \((2, 4)\) is on the circle.