Rough sketch of the circle $x^2+y^2=8x$ and the parabola $y^2=4x$

Question

Find the area lying above x-axis and included between the circle $x^2+y^2=8x$ and the parabola $y^2=4x$.

Once I draw the figure, I know how to apply integration to this. But, if I want to make a rough sketch, how do I know whether the parabola is inscribed in the circle, or the circle is inscribed in the parabola; i.e. which one has more curvature?

Is it possible to predict it from the equations of the curves ?

Can I make use of the concept of eccentricity to find it ?

What I know

$$ x^2-2.4x+16-16+y^2=0\implies (x-4)^2+y^2=4^2 $$ the equation of a circle with radius $4$ and center $(0,4)$

$e_{\text{circle}}=0$ and $e_{\text{parabola}}=1$

Original Plot

Answer 1

One way to find the integration region is to find the interval where one curve lies below the second. In particular, for the circle $$y^2 = 8x-x^2,$$ and for the parabola, $$y^2 = 4x.$$ Solving $$4x \le 8x - x^2$$ results in $$0 \le x \le 4.$$

Answer 2

As you established, for the circle $C$ $$x^2+y^2=8x\to(x-4)^2+y^2=4^2$$which gives you a circle, centre $(4,0)$, radius $4$, which is reasonably easy to sketch

Taking the second equation, you have a hyperbola $H\to y^2=4x$. Taking $y=\pm x$, we have the solutions $x=y=0$ and $x=4, y=\pm 4$

To find which is higher until the intersection points, try $x=2$, then: $$C_{x=2}\to y=\pm\sqrt{12}=\pm2\sqrt3; H_{x=2}\to y=\pm\sqrt8=\pm2\sqrt2$$

Thus note that $C>H$ between $0\le x\le 4$ for $y>0$, and $H>C$ for $y<0$ (the graphs are symmetrical about the $x$-axis)

You can show there are no other intersection points between them, by setting them equal: $$4x=8x-x^2\to x^2-4x=0\to x=0, x=4$$