Question

Find the area of the circle x² + y² = 16 which is exterior the parabola y² = 6x.

Answer 1

The given equations are

x² + y² = 16 …(i)

And y² = 6x …(ii)

On solving the equation (i) and (ii),

Or x² + 6x = 16

Or x² + 6x – 16 = 0

Or (x + 8)(x – 2) = 0

Or x = 2 or – 8 is not possible solution

∴ When x = 2, y = \(\pm \sqrt{6\times 2}=\pm \sqrt{12}\) \(=\pm 2\sqrt{3}\)

Equation (i) is a circle with centre (0, 0) and meets axes at (±4, 0), (0, ±4)

Equation (ii) represents a parabola with axis as x - axis.

Points of intersection are A (2, \(2\sqrt{3}\)) and C (2, \(-2\sqrt{3}\))

These are shown in the graph below:

Area bounded by the circle and parabola

= 2[Area (OADO) + Area (ADBA)]

Area of circle = π(r)²

= π(4)²

= 16π sq. Units

Thus, required area = \(16\pi - \frac{4}{3}[4\pi + \sqrt{3}]\)

The area of the circle x² + y² = 16 which is exterior the parabola y² = 6x is