Area Enclosed By Circle

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 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 = r2

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 area enclosed,

πR2πr2=π(R2r2)=π(R+r)(Rr)\pi R^2 – \pi r^2 = \pi(R^2 – r^2) = \pi(R+r)(R-r)

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:

A=0R2πr  dr=2π×12[r2]0R=πR2A= \int_{0}^{R}2\pi r\; dr = 2\pi \times \frac{1}{2}[r^{2}]^{R}_{0} = \pi R^{2}

 

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 x2 + y2 = a2 and the line joining their points of contact is __________.

Solution:

Problems on Area enclosed by a circle example 1

Equation of chord of contact AB is xh + yk = a2 …..(i)

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

a2h2+k2\frac{a^2}{\sqrt{h^2+k^2}}

Therefore, AB = 2 AM = 2 OA2OM2\sqrt{OA^2 – OM^2} = 2ah2+k2a2h2+k2\frac{2a\sqrt{h^2 + k^2 – a^2}}{\sqrt{h^2+k^2}}

Also, PM = length of perpendicular from P (h, k) to the line (i) is h2+k2a2h2+k2\frac{\sqrt{h^2 + k^2 – a^2}}{\sqrt{h^2+k^2}}

Therefore, the required area of triangle PAB = [1 / 2] * AB * PM = a(h2+k2a2)32(h2+k2)\frac{a (h^2 + k^2 – a^2)^{\frac{3}{2}}}{(h^2+k^2)}.

Example 2: If OA and OB are the tangents from the origin to the circle x2 + y2 + 2gx + 2fy + c = 0 and C is the centre of the circle, the area of the quadrilateral OACB is _________.

Solution:

 

Tangents from the Origin to the Circle

Area of quadrilateral = 2 [area of ΔOAC] =2×12×OA×AC2 \times \frac{1}{2} \times OA \times AC

= S1×g2+f2c\sqrt{S_1} \times \sqrt{g^2+f^2-c}

Point is (0, 0) ⇒ S1 = c.

∴ Area = √c * √[g2 + f2 − c]

Example 3: The equation of the tangent to the circle x2 + y2 = a2 which makes a triangle of area a2 with the co-ordinate axes, is __________.

Solution:

Let the tangent be of form [x / x1] + [y / y1] = 1 and area of Δ formed by it with coordinate axes is [1 / 2] * x1y1 = a2 …….(i)

Again, y1x + x1y − x1y1 = 0

On applying conditions of tangency x1y1x12+y12|-\frac{x_1 y_1}{\sqrt{x_1^2 + y_1^2}}| = a or [ x21+ y21] = [x21 y21] / [a2] ………. (ii)

From (i) and (ii), we get x1 , y1; 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 x2 + y2 = 25 and the co-ordinate axes is ___________.

Solution:

The equation of the tangent at P (3, 4) to the circle x2 + y2 = 25 is 3x + 4y = 25, which meets the co-ordinate 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

⇒ πr2 = 154

⇒r = 7

Hence, the equation of required circle is (x − 1)2 + (y + 1)2 = 72

⇒ x2 + y2 − 2x + 2y = 47.

Example 6: A circle is concentric with the circle x2 + y2 −6x + 12y + 15 = 0 and has area double of its area. The equation of the circle is given by ___________.

Solution:

Equation of circle concentric to given circle is x2 + y2 −6x + 12y + 15 = 0 ……(i)

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

⇒ √[9 + 36 − k] = √2 * √[9 + 36 − 15]

⇒ 45 − k = 60

⇒ k = −15

Hence, the required equation of circle is x2 + y2 −6x + 12y – 15 = 0.

Example 7: Find the smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2.

Solution:

The smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is represented by the shaded area ACBA as

finding the area by integration

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

= 024x2dx02(2x)dx\int_0^2 \sqrt{4-x^2} dx – \int_0^2 (2-x) dx

= [x24x2+42sin1x2]02[2xx22]02[\frac{x}{2} \sqrt{4-x^2} + \frac{4}{2} sin^{-1} \frac{x}{2}]_0^2 – [2x – \frac{x^2}{2}]_0^2

= 2×π2(42)2 \times \frac{\pi}{2} – (4-2)

= (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.

y = a2x2\sqrt{a^2-x^2}

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

y = a2x2\sqrt{a^2-x^2} = a1x2a2a \sqrt{1 – \frac{x^2}{a^2}}

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

(1 / 4) Area of circle = 0aa1x2a2dx\int_0^a a \sqrt{\frac{1-x^2}{a^2}} dx

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

¼ . Area of circle = 0π2a2(1sin2t)cos t dt\int_0^{\frac{\pi}{2}} a^2 (1- sin^2 t) cos\ t\ dt

Now use the trigonometric identity

1sin2t\sqrt{ 1 – sin^2 t } = cos t ; since t varies from 0 to π/2, hence

(1 / 4) Area of circle = 0π2a2cos2t dt \int_0^{\frac{\pi}{2}} a^2 cos^2 t\ dt

Use the trigonometric identity cos2 t = ( cos 2t + 1 ) / 2 to linearise the integrand;

(1 / 4) Area of circle = 0π2a2(cos 2t+1)2dt \int_0^{\frac{\pi}{2}} a^2 \frac{(cos\ 2 t + 1)}{2} dt

Evaluate the integral

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

= (1/4) π a 2

The total area of the circle is obtained through a 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 x2+y2=2x^{2}+y^{2}=2 ​ which is not common to the region bounded by the parabola y2=xy^{2}=x ​ and the straight line y = x, is 

Solution:

Equation of circle = x2+y2=2x^{2}+y^{2}=2 ​

radius = 2\sqrt{2} ​

The required area is

=π×(2)201(xx)dx=2π[23x3212x2] limits from 0 to 1=2π[2312]=2π[16]=12π16=\pi \times (\sqrt{2})^{2}-\int_{0}^{1}(\sqrt{x}-x)dx\\ =2 \pi – [\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{2}x^{2}] \text \ limits \ from \ 0 \ to \ 1 \\ =2 \pi – [\frac{2}{3}-\frac{1}{2}]\\ =2 \pi – [\frac{1}{6}]\\ =\frac{12 \pi – 1}{6} ​

Example 10: What is the area bounded by the circle x2+y2=8x^{2}+y^{2}=8 ​ the parabola x2=2yx^{2}=2y ​ and the line y = x in y ≥ 0?

Solution:

Required area 

=228x2dx2012x2dx02xdx228x2dxApplyTrigSubstitution:x=22sin(u)=π4π48cos2(u)du=8π4π4cos2(u)du=8π4π41+cos(2u)2du=812π4π41+cos(2u)du=812(π4π41du+π4π4cos(2u)du)=812(π2+1)=4(π2+1)=4(π2+1)2012x2dx02xdx=4(π2+1)4302xdx=4(π2+1)432=2π+23=\int _{-2}^2\sqrt{8-x^2}dx-\int _{-2}^0\:\frac{1}{2}x^2dx-\int _0^2\:xdx\\ \int _{-2}^2\sqrt{8-x^2}dx\\\mathrm{Apply\:Trig\:Substitution:}\:x=2\sqrt{2}\sin \left(u\right)\\=\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}8\cos ^2\left(u\right)du\\=8\cdot \int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}\cos ^2\left(u\right)du\\=8\cdot \int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1+\cos \left(2u\right)}{2}du\\=8\cdot \frac{1}{2}\cdot \int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}1+\cos \left(2u\right)du\\=8\cdot \frac{1}{2}\left(\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}1du+\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}\cos \left(2u\right)du\right)\\=8\cdot \frac{1}{2}\left(\frac{\pi }{2}+1\right)\\=4\left(\frac{\pi }{2}+1\right)\\=4\left(\frac{\pi }{2}+1\right)-\int _{-2}^0\frac{1}{2}x^2dx-\int _0^2xdx\\=4\left(\frac{\pi }{2}+1\right)-\frac{4}{3}-\int _0^2xdx\\=4\left(\frac{\pi }{2}+1\right)-\frac{4}{3}-2\\=2\pi +\frac{2}{3} ​

Example 11: What is the area bounded by the circles x2+y2=r2x^{2}+y^{2}=r^{2}, r = 1, 2 and the rays 2(x)23xy2(y)2=02(x)^{2}-3xy-2(y)^{2}=0?

Solution: 

2(x)23xy2(y)2=02(x)24xy+xy2(y)2=02x(x2y)+y(x2y)=0(2x+y)(x2y)=0y=2x or y=x22(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 \text \ or \ y=\frac{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=x2y=-2x \text \ and \ y=\frac{x}{2}. The slope of the line y = -2x is negative and passes through the origin. Similarly, the slope of the line y=x2y=\frac{x}{2} is found. 

Hence, the slope of the line y = -2x is m = -2 and y=x2y=\frac{x}{2} is -1 / 2.

m1×m2=1m_{1} \times m_{2}=-1

So, the lines are perpendicular.

The required area is

A=π×224π×124=ππ4=3π4A=\frac{\pi \times {2}^{2}}{4}-\frac{\pi \times {1}^{2}}{4}\\ =\pi-\frac{\pi}{4}\\ =\frac{3\pi}{4}