
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.
To solve the problem of identifying the point lying on the smallest circle that touches the parabola \( y = 6 - x^2 \) and the lines \( y = \sqrt{3} |x| \), we need to analyze the geometric positioning and the constraint on the circle.
After analyzing the options, it turns out that the point \( (2, 4) \) satisfies the tangency condition for the circle touching both described curves in the context of enclosing it within the arcs of the parabola and the lines.
Hence, the correct answer is \( (2, 4) \).
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16
Let the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
Match the LIST-I with LIST-II for an isothermal process of an ideal gas system. 
Choose the correct answer from the options given below:
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?
