Q.To reduce traffic and avoid red lights, roundabouts are often built on busy roads. Such a roundabout is built such that its boundary $C_1 : x^2 + y^2 = 64$ is given. In the middle of this roundabout, there is a circular pond with a fountain, whose equation is $C_2 : x^2 + y^2 = 4$. Based on the above information, answer the following questions:

Part (a)

(i) $C_1:x^2+y^2=64$ and $C_2:x^2+y^2=4$ are concentric circles (centre $O$) of radii $8$ and $2$; $C_2$ lies inside $C_1$.

(ii) $C_1:y=\pm\sqrt{64-x^2}$; $C_2:y=\pm\sqrt{4-x^2}$.

(iii)(A) Area of the roundabout ($C_1$): …

For a circle $x^2+y^2=r^2$, integrating the upper half $y=\sqrt{r^2-x^2}$ over $[-r,r]$ and doubling recovers the area $\pi r^2$; equivalently, four times the first-quadrant integral.

Part (a)

(i) $C_1:x^2+y^2=64$ is a circle centred at the origin with radius $8$ (the roundabout boundary); $C_2:x^2+y^2=4$ is concentric with radius $2$ (the pond). Draw two concentric circles, the smaller inside the larger.

(ii) Solving for $y$: $C_1:y=\pm\sqrt{64-x^2}$ and $C_2:y=\pm\sqrt{4-x^2}$.

(iii)(A) Using symmetry (four equal quadrants) and $\int\sqrt{a^2-x^2}dx=\tfrac x2\sqrt{a^2-x^2}+\tfrac{a^2}{2}\sin^{-1}\tfrac xa$ with $a=8$: …

Method: Area of a circle (and a ring between two circles) by integration

Use this whenever a region is enclosed by a circle $x^2 + y^2 = r^2$, or is the ring left between an outer and an inner concentric circle.

Steps

Step 1: Read each radius.

From $x^2 + y^2 = r^2$ the radius is $r = \sqrt{r^2}$. If two circles are given, name the larger radius $R$ and the smaller $r$.

Step 2: Set up one circle's area with symmetry.

A circle is symmetric about both axes, so integrate the first-quadrant arc $y = \sqrt{r^2 - x^2}$ and multiply by $4$:

using

$\int \sqrt{r^2 - x^2}\,dx = \frac{x}{2}\sqrt{r^2 - x^2} + \frac{r^2}{2}\sin^{-1}\frac{x}{r} + C.$ …

Common Mistakes

Mistake 1: Misreading the radius from $x^2+y^2=64$

Students take the radius as $64$ instead of $8$. Why it's wrong: the standard form is $x^2+y^2=r^2$, so $r^2=64$ gives $r=8$ (and $r^2=4$ gives $r=2$). Correct approach: always square-root the constant on the right.

Mistake 2: Subtracting the pond when the question asks for the whole roundabout

Part (iii)(A) asks for the area enclosed by the entire roundabout $C_1$, which is the full disc $64\pi$ — not the ring between the two circles. Why it's wrong: subtracting the pond ($64\pi-4\pi=60\pi$) answers a different question. Correct approach: read carefully — the full roundabout is $64\pi$, the pond alone is $4\pi$. …