# Find the area common to the circle x^2 + y^2 = 16 a^2 and the parabola y^2 = 6ax.

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Find the area common to the circle x2 + y2 = 16 a2 and the parabola y2 = 6ax.

OR

Find the area of the region {(x, y) : y2 ≤ 6ax} and {(x, y) : x2 + y2 ≤ 16a2}.

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To find area given equations are

y2 = 6ax ...(i)

x2 + y2 = 16a2 ...(ii)

On solving Equation (i) and (ii)

Or x2 + (6ax)2 = 16a2

Or x2 + (6ax)2 – 16a2 = 0

Or (x + 8a) (x – 2a) = 0

Or x = 2a or x = – 8a is not possible solution.

Then y2 = 6a(2a) = 12a2 = $\pm$2√3a

Equation (i) represents a parabola with vertex (0, 0) and axis as x - axis.

Equation (ii) represents a with centre (0, 0) and meets axes (±4a, 0), (0, ±4a).

Point of intersection of circle and parabola are (2a, 2√3a), (2a, – 2√3a).

These are shown in the graph below: -

Required area = 2[Region ODCO + Region BCDB]

The area common to the circle x2 + y2 = 16a2 and the parabola y2 = 6ax is $\frac{4a^2}{3}(4\pi + \sqrt{3})$ sq. units