Question

If the area enclosed between the circle \(x^2+y^2=25\) and the parabola \(y^2=16x\) is \(A\), then \(A=\ ?\)

• A. \(25\pi-96\)
• B. \(50\pi-96\)
• C. \(25\pi-48\)
• D. \(50\pi-48\)

Hint:

Whenever a circle and parabola intersect symmetrically about the \(x\)-axis, integrate only upper half and multiply by 2 to reduce calculation.

Answer 1

The Correct Option is A

Concept: Area between curves is obtained by: \[ \int (\text{upper curve}-\text{lower curve})dx \] or by subtracting known geometric regions. The given curves are: \[ x^2+y^2=25 \] which is a circle of radius \(5\), and \[ y^2=16x \] which is a parabola opening rightwards.

Step 1: Finding points of intersection.
From parabola: \[ x=\frac{y^2}{16} \] Substitute into circle: \[ \left(\frac{y^2}{16}\right)^2+y^2=25 \] \[ \frac{y^4}{256}+y^2=25 \] Multiply by 256: \[ y^4+256y^2-6400=0 \] Let: \[ y^2=t \] Then: \[ t^2+256t-6400=0 \] Solving: \[ (t-16)(t+400)=0 \] Thus: \[ t=16 \] Hence: \[ y=\pm4 \] Corresponding \(x\): \[ x=\frac{16}{16}=1 \] So points of intersection are: \[ (1,4)\quad \text{and}\quad (1,-4) \]

Step 2: Finding required area.
Area inside circle: \[ =25\pi \] Area cut by parabola: \[ A_1=2\int_0^4 \frac{y^2}{16}\,dy \] \[ =\frac18\int_0^4 y^2\,dy \] \[ =\frac18\left[\frac{y^3}{3}\right]_0^4 \] \[ =\frac18\cdot\frac{64}{3} \] \[ =\frac83 \] Area of corresponding strip in circle: \[ A_2=2\int_0^4 \sqrt{25-y^2}\,dy \] Using standard formula: \[ \int \sqrt{a^2-y^2}\,dy = \frac y2\sqrt{a^2-y^2} +\frac{a^2}{2}\sin^{-1}\frac ya \] After evaluation: \[ A_2=25\pi-\frac{280}{3} \] Thus enclosed area: \[ A=A_2-A_1 \] \[ =25\pi-\frac{280}{3}-\frac83 \] \[ =25\pi-96 \] Hence: \[ \boxed{25\pi-96} \]