How to Calculate the Area Enclosed by a Circle

A circle is the locus of all points in a plane at a fixed distance called the radius from a fixed centre. The space covered inside the boundary of a circle is referred to as the area enclosed by the circle.

Equation and Standard Formulation for Area Enclosed by $x^2 + y^2 = a^2$

The equation $x^2 + y^2 = a^2$ represents a circle with centre at the origin $(0, 0)$ and radius $a$ units. Every point $(x, y)$ that satisfies this relation lies exactly $a$ units from the origin, thus forming the complete locus for the circle.

The general formula for the area enclosed by a circle of radius $r$ is given by:

$\text{Area} = \pi r^2$

For the circle $x^2 + y^2 = a^2$, the radius is $a$. Therefore, the area enclosed is $\pi a^2$.

Deduction of Circle Area Using Sector Rearrangement

A rigorous geometric argument considers dividing the circle of radius $r$ into $n$ congruent sectors, where $n$ is large. Each sector approaches the shape of a triangle with base $\approx r \, d\theta$ and height $r$. When these sectors are rearranged alternately, they nearly form a rectangle whose dimensions stabilize as $n \to \infty$.

The ‘rectangle’ thus formed has approximate length $\pi r$ (half the circumference; sum of alternate arcs) and width $r$. The area of the rectangle is then $\pi r \cdot r = \pi r^2$, confirming the area formula for a circle.

Explicit Calculation of Area Enclosed by $x^2 + y^2 = a^2$ Using Definite Integration

To rigorously compute the area, consider the circle $x^2 + y^2 = a^2$. By symmetry, it suffices to evaluate the area in the first quadrant and multiply by $4$.

For $x \in [0, a]$, the upper semicircle is described by $y = \sqrt{a^2 - x^2}$. The area in the first quadrant, $A_1$, is given by integrating $y$ with respect to $x$ from $x = 0$ to $x = a$:

$A_1 = \displaystyle\int_{0}^{a} \sqrt{a^2 - x^2}\, dx$

The total area is $A = 4A_1 = 4 \displaystyle\int_{0}^{a} \sqrt{a^2 - x^2}\, dx$.

Let us compute the integral $\displaystyle\int_{0}^{a} \sqrt{a^2 - x^2}\, dx$ using trigonometric substitution. Set $x = a\sin\theta \implies dx = a\cos\theta\, d\theta$, with $\theta$ varying from $0$ (when $x = 0$) to $\pi/2$ (when $x = a$):

$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta$

Therefore, the integral becomes:

$\displaystyle\int_{0}^{a} \sqrt{a^2 - x^2}\, dx = \int_{0}^{\pi/2} a\cos\theta \cdot a\cos\theta\, d\theta = a^2 \int_{0}^{\pi/2} \cos^2 \theta \, d\theta$

Recall that $\cos^2 \theta = \frac{1+\cos(2\theta)}{2}$. Thus,

$a^2 \int_{0}^{\pi/2} \frac{1+\cos(2\theta)}{2} d\theta = \frac{a^2}{2} \int_{0}^{\pi/2} 1\, d\theta + \frac{a^2}{2} \int_{0}^{\pi/2} \cos(2\theta)\, d\theta$

$\frac{a^2}{2} [\theta]_{0}^{\pi/2} + \frac{a^2}{2} \left[ \frac{\sin(2\theta)}{2} \right]_{0}^{\pi/2}$

The first term: $[\theta]_{0}^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

The second term: $\left[ \frac{\sin(2\theta)}{2} \right]_{0}^{\pi/2} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0 - 0 = 0$

Thus, the value is $\frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{a^2 \pi}{4}$.

Hence, $A_1 = \frac{a^2 \pi}{4}$ and the total area is $A = 4 \times \frac{a^2 \pi}{4} = \pi a^2$.

Area Enclosed by Other Circles: Examples for $x^2 + y^2 = k$

For a circle given by $x^2 + y^2 = k$, the radius is $r = \sqrt{k}$, as $\sqrt{x^2 + y^2} = r$. The area enclosed is therefore $\pi k$.

For example, for $x^2 + y^2 = 4$, the radius is $2$, and the area is $\pi \times (2)^2 = 4\pi$.

For $x^2 + y^2 = 8$, the radius is $2\sqrt{2}$, and the area is $\pi (2\sqrt{2})^2 = \pi \times 8 = 8\pi$.

Worked Examples on Area Enclosed by a Circle

Example 1: For the circle $x^2 + y^2 = 16$, compute the area enclosed.

Given $x^2 + y^2 = 16$, the radius is $r = \sqrt{16} = 4$.

Area = $\pi r^2 = \pi \times 4^2 = 16\pi$

Example 2: Find the area enclosed by the circle $x^2 + y^2 = 2$.

Here, the radius is $r = \sqrt{2}$.

Area = $\pi r^2 = \pi \times (\sqrt{2})^2 = 2\pi$

Example 3: If the circumference of a circle is $C = 40\ \mathrm{cm}$, calculate the area enclosed.

Given circumference $C = 2\pi r = 40$

$r = \dfrac{40}{2\pi} = \dfrac{20}{\pi}$

Area = $\pi r^2 = \pi \left(\dfrac{20}{\pi}\right)^2 = \pi \dfrac{400}{\pi^2} = \dfrac{400}{\pi}$

Example 4: For the circle $x^2 + y^2 = 25$, compute the area enclosed and compare with [Area Of A Circle Formula].

Here, radius $r = \sqrt{25} = 5$

Area = $\pi r^2 = \pi \times 25 = 25\pi$

Key Remarks on Definition, Area, and Application

If a circle is given in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, the centre is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$. The area enclosed is then $\pi (g^2 + f^2 - c)$. For all such cases, the explicit derivation above applies once the correct value of radius is identified.

To deepen understanding, refer also to [Area Of Hexagon Formula], where methods for polygonal regions are given for comparison.

Area Enclosed by a Circle Cut by a Line or Chord (Sector and Segment)

If a circle is intersected by a straight line (a chord or secant), the region enclosed between the chord and the arc is called a segment. The area of a sector subtended by central angle $\theta$ (in radians) in a circle of radius $r$ is $A_\text{sector} = \dfrac12 r^2 \theta$. The area of the segment is $A_\text{segment} = A_\text{sector} - A_\text{triangle}$, where $A_\text{triangle}$ represents the area of the triangle formed by the two radii and the chord.

For related results on areas involving other figures, reference [Area Of Triangle Formula].

Frequently Asked Questions on Area Enclosed by a Circle

Question: What is the unit of area enclosed by a circle?

The area is measured in square units corresponding to the unit of radius (e.g., $\text{cm}^2$, $\text{m}^2$).

Question: What happens to the area if the radius doubles?

If the radius doubles ($r$ becomes $2r$), the area increases by a factor of $4$, as area is proportional to $r^2$.

For more on composite and related geometric areas, see [Area And Perimeter Formula].