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.