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 \( C \)?

Updated On: Nov 27, 2024
  • \( (2, 4) \)
  • \( (1, 2) \)
  • \( (2, 2) \)
  • \( (1, 1) \)
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The Correct Option is A

Solution and Explanation

Sol. Fig
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.

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