A locus of a point whose distance is constant from the given point is a circle. The centre of the circle is a fixed point, and the fixed distance is the radius. The circumference is the perimeter of the circle. This article explains how to find the **area enclosed by circle**.

The equation of a circle with radius r and centre (a, b) is given by

**(x + a) ^{2 }+ (y + b)^{2} = r^{2}**

## The Formula for Area Bounded by a Circle

The area enclosed by 2 concentric circles is given by

- If R and r are radii of two concentric circles, then the area enclosed is

The method of integration is used to find the areas and volumes by adding up small pieces. A series of concentric rings with radius r where r varies from 0 at the origin to R outside the circle.

2πr gives the circumference, which is the area of each ring.

The formula for finding the area bounded by a circle through integration is given by

## Problems on Area Enclosed by a Circle

**Example 1:** The area of the triangle formed by the tangents from the points (h, k) to the circle x^{2} + y^{2} = a^{2} and the line joining their points of contact is __________.

**Solution:**

Equation of chord of contact AB is xh + yk = a^{2} …..(i)

OM = Length of the perpendicular from O (0, 0) on line (i)

Therefore, AB = 2 AM

Also, PM = Length of the perpendicular from P (h, k) to the line (i)

Therefore, the required area of triangle PAB = [1 / 2] * AB * PM

**Example 2:** If OA and OB are the tangents from the origin to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 and C is the centre of the circle, the area of the quadrilateral OACB is _________.

**Solution:**

Area of quadrilateral = 2 [area of ΔOAC]

Point is (0, 0) ⇒ S_{1 }= c.

∴ Area = √c * √[g^{2} + f^{2} − c]

**Example 3:** The equation of the tangent to the circle x^{2} + y^{2} = a^{2}, which makes a triangle of area a^{2} with the co-ordinate axes, is __________.

**Solution:**

Let the tangent be of the form [x / x_{1}] + [y / y_{1}] = 1, and the area of Δ formed by it with coordinate axes is

_{1}y

_{1}= a

^{2}…….(i)

Again, y_{1}x + x_{1}y − x_{1}y_{1} = 0

On applying conditions of tangency

or

[ x^{2}

_{1}+ y

^{2}

_{1}] = [x

^{2}

_{1}y

^{2}

_{1}] / [a

^{2}] ………. (ii)

From (i) and (ii), we get x_{1 and} y_{1,} which gives the equation of tangent as x ± y = ± a√2.

Trick: There may be 4 tangents. The lines x ± y = ± a√2 make a triangle of the area in all four quadrants.

**Example 4:** The area of the triangle formed by the tangent at (3, 4) to the circle x^{2} + y^{2} = 25, and the co-ordinate axes is ___________.

**Solution: **

The equation of the tangent at P (3, 4) to the circle x^{2} + y^{2} = 25 is 3x + 4y = 25, which meets the coordinate axes at A (25/3, 0) and B (0, 25/4).

If O is the origin, then the ΔOAB is a right-angled triangle with OA = 25/3 and OB = 25/4.

Area of the ΔOAB = (1/2) × OA × OB

= (1/2) × (25/3) × (25/4)

= 625 / 24

**Example 5:** The lines 2x − 3y = 5 and 3x − 4y = 7 are the diameters of a circle of area 154 square units. The equation of the circle is _________.

**Solution:**

Centre of circle = Point of intersection of diameters = (1, 1)

Now, area = 154

⇒ πr^{2} = 154

⇒ r = 7

Hence, the equation of required circle is (x − 1)^{2} + (y + 1)^{2} = 7^{2}

⇒ x^{2} + y^{2} − 2x + 2y = 47.

**Example 6:** A circle is concentric with the circle x^{2} + y^{2} −6x + 12y + 15 = 0 and has an area double its area. The equation of the circle is given by ___________.

**Solution:**

Equation of circle concentric to given circle is x^{2} + y^{2} −6x + 12y + 15 = 0 ……(i)

The radius of the circle (i) = √2 (Radius of the given circle)

⇒ √[9 + 36 − k] = √2 * √[9 + 36 − 15]

⇒ 45 − k = 60

⇒ k = −15

Hence, the required equation of the circle is x^{2} + y^{2} −6x + 12y – 15 = 0.

**Example 7:** Find the smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2.

**Solution:**

The smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is represented by the shaded area ACBA as

It can be observed that Area ACBA = Area OACBO – Area (ΔOAB)

= (π – 2) Units

**Example 8: **Find the area of a circle with radius a.

**Solution: **

The equation of the circle is given by x^{ 2} + y^{ 2} = a^{ 2}

The circle being symmetric about the x and y axes, the area of the circle can be found by multiplying 4 in order to arrive at the total area of the circle.

Solve the above equation for y.

The equation of the upper semi-circle (y positive) is given by

Use integrals to find the area of the upper right quarter of the circle as follows:

Substitute x / a by sin t so that sin t = x / a and dx = a cos t dt, and the area is given by

Now, use the trigonometric identity.

Use the trigonometric identity cos^{2} t = ( cos 2t + 1 ) / 2 to linearise the integrand;

Evaluate the integral.

(1 / 4) Area of circle = (1/2) a^{ 2} [ (1/2) sin 2t + t ]_{0}^{π/2}

= (1/4) π a^{ 2}

The total area of the circle is obtained through multiplication by 4.

Area of circle = 4 * (1/4) π a^{ 2} = π a^{ 2}

**Example 9:** The area of the region, enclosed by the circle x^{2} + y^{2} = 2, which is not common to the region bounded by the parabola y^{2} = x and the straight line y = x, is

**Solution:**

The equation of the circle is: x^{2} + y^{2} = 2

Radius = √2

The required area is

**Example 10: **What is the area bounded by the circle x^{2} + y^{2} = 8, the parabola x^{2} = 2y and the line y = x in y ≥ 0?

**Solution:**

Required area

= ∫_{-2}^{2} √(8-x^{2}) dx – ∫_{-2}^{0} ½ x^{2} dx – ∫_{0}^{2} x dx

= 2∫_{0}^{2}√(8-x^{2}) dx – [x^{3}/6]_{-2}^{0} – [x^{2}/2]_{0}^{2}

= 2[(x/2)√(8-x^{2}) + (8/2)sin^{-1}(x/2√2)]_{0}^{2} – (4/3) – 2

= 2[2 + 4.π/4] – 10/3

= (⅔) + 2π

**Example 11: **What is the area bounded by the circles x ^{2 }+ y^{2} = r^{2}, r = 1, 2 and the rays 2(x)^{2} – 3xy – 2(y)^{2} = 0?

**Solution: **

2(x)^{2} – 3xy – 2(y)^{2} = 0

2(x)^{2} – 4xy + xy – 2(y)^{2} = 0

2x(x – 2y) + y(x – 2y) = 0

(2x + y) (x – 2y) = 0

y = -2x or y = x/2

When the two circles of radius 1 and 2 are drawn, the circle with radius 2 is bigger than the circle with radius 1. Now, we have two lines, y = -2x and y = x/2. The slope of the line y = -2x is negative and passes through the origin. Similarly, the slope of the line y = x/2 is found.

Hence, the slope of the line y = -2x is m = -2, and y = x/2 is -1 / 2.

m_{1}m_{2} = -1

So, the lines are perpendicular.

The required area is

A = π(2^{2})/4 – π(1)^{2}/4

= π – π/4

= 3π/4

## Frequently Asked Questions

### Give the formula for the area enclosed by two concentric circles.

If R and r are radii of two concentric circles, then the area enclosed is given by the formula π(R^{2 }– r^{2}).

### What is the area enclosed by a circle of radius r?

Area of the circle = πr^{2}, where r is the radius.

### What is the difference between the circumference and the area of a circle?

The space enclosed by a circle is called the area. The length of the curved surface is called the circumference of the circle. The area of the circle is πr^{2}, whereas the circumference is given by 2πr (radius is r).

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