We have learned what is a circle, its properties, and how to find its area and circumference earlier on another page. Let us recall, what is a circle. A collection of points in a plane equidistant from a fixed point is called a circle. The fixed point is called the center and the equidistant distance is called the radius of the circle.

Now let us study concentric circles. On this page we will be learning what is concentric circle theory? Its properties, concentric circle’s equation, and concentric circles examples.

Concentric circle theory is a very vital concept in geometry.

Two or more circles sharing the same center but different radii are called concentric circles.

From the figure, we can say that two circles are given with the same center and radius r for smaller circles and R for the larger circles. They are called concentric circles.

And the region between the circles is called an annulus. From the above figure, you can say that the blue-colored part is called an annulus. An example of the annulus is a ring, circular road, etc.

Let us derive the concentric circles’ equation. Concentric circles equation will help you in solving example questions.

Given that two concentric circles with the same center and different radii.

Let us consider a circle with center (-g, -f) and radius \[\sqrt {({g^2} + {f^2} - c)} \]

Therefore, the equation for the given circle will be x2 + y2 + 2gx + 2fy + c =0

As both, the circles have same center (-g, -f), but they have different radii, c ≠ c’

Therefore, the equation of the other circle will be x2 + y2 + 2gx + 2fy + c’ =0

Similarly, a circle with center (h, k), and the radius is equal to r, then the equation becomes

( x – h )2 + ( y – k )2 = r2

Therefore, the concentric circles equation is ( x – h )2 + ( y – k )2 = r12, Where r ≠ r1

By assigning different values to the radius in the above equation, we shall get a number of circles.

We can understand this concept by solving concentric circles examples.

The region between the two concentric circles is called the annulus. It can be assumed as a flat ring.

From the above figure, two circles are given, where a small circle lies inside the bigger one. The point O is the center of both circles. The shaded colored region, between these two circles, is known as an annulus. The smaller circle is known as the inner circle, while the bigger circle is termed as the outer circle. Now let us derive the formula for the area of annulus.

We can calculate the area of the annulus by calculating the area of both the circles and subtracting them.

In the above figure, two circles are given common center O. Let R be the radius of the outer circle, and r be the radius of the inner circle. The shaded portion represents an annulus. To find the area of this annulus, we have to find the areas of both circles.

Therefore,

Area of Outer Circle with radius R = πR2

Area of Inner Circle with radius r = πr2

Now, Area of Annulus = Area of Outer Circle(with radius R) – Area of Inner Circle(with radius r)

Now let us solve some concentric circles examples.

Q.1: Calculate the area of an annulus whose outer radius is 21 cm and the inner radius 14 cm?

Solution: Given that outer radius R = 21 cm

inner radius r = 14 cm

Area of outer circle = πR2

= \[\frac{{22}}{7}\] x 21 x 21

= 22 x 21 x 3

= 1386cm2

Area of inner circle = πr2

= \[\frac{{22}}{7}\] x 14 x 14

= 22 x 14 x 2

=616cm2

Area of the annulus = Area of the outer circle – Area of the inner circle

Area of the annulus = 1386 – 616

Area of the annulus = 770cm2

Example 2: Equation of circle passing through (-1,-2) and concentric with the circle x2 +y2 +3x + 4y +1 =0 ?

Solution: A circle with center ( a, b) and radius r

We have ( x- a)2 + ( y - b)2 = r2

now,

x2 + y2 + 3x + 4y + 1 = 0

(x + \[\frac{3}{2}\])2 + ( y + 2)2 - 4 - \[\frac{9}{4}\] + 1 = 0

(x + \[\frac{3}{2}\])2 + ( y + 2)2 - \[\frac{{21}}{4}\] = 0

(x + \[\frac{3}{2}\])2 + ( y + 2)2 = \[\frac{{21}}{4}\]

The given circle has the center ( -\[\frac{3}{2}\], -2) and radius\[\sqrt {\frac{{21}}{4}} \]

Therefore our circle equation will be

(x + \[\frac{3}{2}\])2 + ( y + 2)2 = r2

But it is passing through ( -1, -2)

( -1 + \[\frac{3}{2}\])2 + ( -2 + 2)2 = r2

\[{\left( {\frac{1}{2}} \right)^2}\]= r2

r = \[\frac{1}{4}\]

Therefore the equation is (x + \[\frac{3}{2}\])2 + ( y + 2)2 =\[\frac{1}{4}\]

Some more concentric circles examples to solve.

Q 1. Find the equation of the circle concentric with the circle with the equation x2 + y2 + 4x – 8y – 12 =0, having the radius double of its radius.

Q 2.Find the equation of the circle which is concentric with the circle 2x2 +2y2 + 3x + 4y + 5 whose radius is \[3\sqrt 5 \]

FAQ (Frequently Asked Questions)

1. What is a circle inside a triangle called?

A circle inside a triangle is called the incircle or inscribed circle of a triangle. It is the largest circle in the triangle and it touches all the three sides of a triangle.

The center of the incircle is called the incenter of the circle.

2. What are the properties of a circle?

Some of the properties of the circle are as follows:

Circles with the same radii are said to be congruent.

The longest chord of a circle is the diameter of a circle.

The radius that is drawn perpendicular to the chord bisects the chord

A circle can be inscribed in a square, triangle or a kite

The chords that are equidistant from the center are equal in length

The distance from the center of the circle to the longest chord (diameter) is zero

The perpendicular distance from the center of the circle decreases when the length of the chord increases

An isosceles triangle is formed when the radii joining the ends of a chord to the center of a circle