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As we already know that a circle is a geometric figure formed by a collection of points that are equidistant from a fixed point.The fixed point is known as the center and the equidistant distance is called the radius of the circle. In geometry, when objects share a common center, they are said to be concentric. Examples of concentric objects are circles, spheres, regular polyhedra, regular polygons, etc., as they share a common center point. According to the Euclidean Geometry, two concentric circles should have different radii from each other.
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Key Terms: Concentric Circles, Annulus, Euclidean Geometry, chord, radius, center point.
What are Concentric Circles?
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Concentric circles are two or more circles having a common center point but different radii. The region between the two concentric circles is known as the annulus.
Two concentric circles can never intersect at any point and the region and distance of the annulus between the circles will always be the same. Some examples of annulus include a ring, circular road, etc. A dartboard also has concentric circles and these circles form around a common center point called the bull’s eye.

Concentric Circles
Read more: Formation of a Differential Equation
Equations for Concentric Circles
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Let the equation of a circle with center (-g, -f) and radius [g2+f2-c] be
x2 + y2 + 2gx + 2fy + c =0.
Therefore, the equation of the circle concentric with the other circle will be
x2 + y2 + 2gx + 2fy + c’ =0.
Similarly, a circle with the center (h, k), and the radius equal to r, then the equation becomes
( x – h )2 + ( y – k )2 = r2
Therefore, the equation of a circle concentric with the circle is written as:
(x – h)2 + (y – k)2 = r12, where r ≠ r1
We can obtain a family of circles by putting different values to the radius mentioned in the equation.
Thus, it can be observed that, both the equations have the same center (-g, -f), but they have different radii, where c≠ c’ .
Read more: Bayes Theorem Formula
Concentric Circles Theorem
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Theorem - In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Given- Two concentric circles C1 and C2, with center O and a chord AB of the larger circle C1, touching the smaller circle C2 at the point P .
To Prove- AP = BP
Proof-
- Join OP.
- Since AB is the chord of larger circle C1, it becomes the tangent to C2 at P. And, OP is the radius of circle C2.
- Now, we know that the radius is perpendicular to the tangent at the point of contact.
Therefore, OP ⊥ AB .
- Similarly, AB is a chord of the circle C1 and OP ⊥ AB.
Therefore, OP is the bisector of the chord AB.
- Thus, the perpendicular from the center bisects the chord, i.e., AP = BP.
Hence, PROVED.
Read more: Surface Area of a Cone Formula
Annulus- The Region Between Concentric Circles
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The region formed between two concentric circles is called the annulus. It is shaped like a flat ring. We can calculate the area and perimeter of an annulus by using some formulas. The area of the annulus is calculated by subtracting the area of smaller circles from the area of the larger circle.
Thus, the area of Annulus can be calculated by
Area of annulus = πR2 – πr2
Where, R is the radius of the larger circle and r is the radius of the smaller circle.
Things to Remember
- For two circles to be concentric, they must have a common center point but different radii.
- Two concentric circles never intersect at any point.
- The region of the annulus, i.e., The distance between the two circles is equal all the way.
- The area of the annulus is calculated by subtracting the area of smaller circles from the area of the larger circle.
- Area of annulus = πR2 – πr2
Sample Questions
Ques. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. (5 marks)
Ans. Given: Let two concentric circles be C1 & C2 with center O.
AB be chord of the larger circle which touches the smaller circle C1 at point P.
To find: Length of AB
Solution: Connecting OP, OA and OB
OP= Radius of smaller circle = 3 cm
OA= OB = Radius of larger circle = 5 cm
Since AB is tangent to circle
OP is perpendicular to AB.
(Tangent at any point of circle is perpendicular to the radius through point of contact)
Therefore, angle OPA = angle OPB = 90°.
Thus, in right triangle OAP,
OA2 = OP2 + AP2
This implies, 52=32+AP2
This implies, AP=4cm
Similarly, in right triangle OPB,
OB2 =OP2 + PB2
This implies, 52= 32+ PB2
Therefore, PB=4cm.
Hence, AB=AP+PB =4+4=8cm.
Ques. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If angle AOB = 30°, find the area of the shaded region. (4 marks)
Ans. Area of shaded region = Area of sector AOB - Area of sector COD
Area of sector AOB
radius = r = 21 cm and angle = 30°
Therefore, area of sector AOB = angle/360 x π.r2
=30/360 x 22/7 x 21 x 21
=231/2 cm2
Area of sector COD
radius = r = 7 cm and angle = 30°.
=angle/360 x π.r2
==30/360 x 22/7 x (7)2 = 77/6cm2
Therefore, Now,
Area of shaded region = Area of Sector OAB - Area of sector OCD
=231/2 - 77/6
=308/3 cm2
Ques. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD. (4 marks)
Ans. In circle,
OP perpendicular to BC
(As OP is perpendicular to line/)
So, OP bisects BC,
(Perpendicular drawn from centre of a circle to a chord)
i.e., BP = CP (1)
In circle C2
OP is perpendicular to AD
(As OP is perpendicular to line /)
So, OP bisects AD,
i.e. AP= DP (2)
(Perpendicular drawn from centre of a circle to a chord bisects the chord).
Subtracting (2) and (1),
AP-BP=DP-CP
This implies, AB=CD [Hence, Proved]
Ques. Calculate the area of an annulus whose outer radius is 21 cm and the inner radius 14 cm. (2 marks)
Ans. Given that outer radius R = 21 cm
inner radius r = 14 cm
Area of outer circle = πR2 = 22/7 x 21x 21 =1386cm2
Area of inner circle = πr2 = 22/7 x 14 x 14 =616cm2
Area of the annulus = Area of the outer circle – Area of the inner circle
Therefore, Area of the annulus = 1386 – 616 Area of the annulus = 770cm2
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