Q. 375.0( 3 Votes )

# Find the area of the circle x^{2} + y^{2} = 16 which is exterior the parabola y^{2} = 6x.

Answer :

The given equations are

x^{2} + y^{2} = 16 …(i)

And y^{2} = 6x …(ii)

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

Or x^{2} + 6x = 16

Or x^{2} + 6x – 16 = 0

Or (x + 8)(x – 2) = 0

Or x = 2 or – 8 is not possible solution

∴ When x = 2, y = ± = ± = ±

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 () and C (2,)

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

The area of the circle x^{2} + y^{2} = 16 which is exterior the parabola y^{2} = 6x is

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