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Circles Connected With Triangle

Solution of Triangles is one of the fundamental applications of trigonometry. And when we deal with solution of triangles, we have various circles associated with triangles. These circles and the related trigonometric formulae are extremely helpful in solving questions related to triangles. We begin with what exactly do we mean by a circle and then discuss the various types of circles associated with triangles.

A circle is made up of an infinite set of points which are all equidistant from a point called the center.
There are basically three types of circles associated with the triangles.

Circum-Circle of a triangle

Let ABC be any triangle. A circle passing through all three vertices of the triangle is called the circum-circle of the triangle. The center of the circumcircle is called as the circumcentre of the triangle. Circumcentre is also the point of intersection of perpendicular bisectors of the sides of the triangle.

The radius of this circumcircle is termed as the circum radius of the triangle and is denoted by R.

In the triangle ABC, the circumcentre R is given by 

R = a/2 sin A = b/2 sin B = c/2 sin C = abc/4Δ, where

Δ is the area of the triangle and s = (a+b+c)/2.


For more on circum-circle, please refer the video


In-circle of a triangle

Contrary to the circum-circle, if a circle is constructed inside the triangle in such a way that the circle touches all the sides of the triangle is called an in-circle. It is also called an inscribed circle.

The centre of the incircle is called an in centre of the triangle. The incentre is also the point of intersection of internal bisectors of the triangle. The radius of the in circle is called the in radius of triangle and is denoted by ‘r’. The various trigonometric formulae for evaluating the in radius are listed below:

  • r = Δ/s 

  • r = (s – a) tan A/2 = (s – b) tan B/2 = (s – c) tan C/2 

  • r = 4R sin (A/2) sin (B/2) sin (C/2)

  • r = [a sin (B/2) sin (C/2)]/ cos (A/2)

 = [b sin (A/2) sin (C/2)]/ cos (B/2)

 = [c sin (B/2) sin (A/2)]/ cos (C/2)


Escribed circle

The circle which touches one of the sides of the triangle and the rest two sides when produced is called an escribed circle. Hence, an escribed circle is in the exterior of the triangle.  


In case of escribed circle, the circle acts as a tangent to the side AC of the triangle and to the sides AB and BC on being extended. In this case  the escribed circle is opposite angle B.


In the figure given below, there is a circle escribed opposite to triangle ABC. r1 is the radius of the circle and O is the center. Then,

r1 = Δ/(s – a) = s tan A/2 = 4R sin A/2 cos B/2 cos C/2,

r2 = Δ/(s – b) = s tan B/2 = 4R cos A/2 sin B/2 cos A/2,

r3 = Δ/(s – c) = s tan C/2 = 4R sin A/2 cos A/2 cos B/2,


A figure depicting escribed circles opposite to every side of the triangle is given below


Some other related concepts of trigonometry:

The Orthocenter of a Triangle: 

The orthocenter of a triangle is the point at which all the three altitudes drawn on the opposite sides of a triangle from the vertices meet.


Pedal Triangle:

By a pedal triangle we mean a triangle obtained by joining the end points of altitudes on the three sides. Here LMN is the pedal triangle obtained by joining the feet of the perpendiculars on the three sides. 


  • The circum radius of the pedal triangle is given by = R/2.

  •  The area of the pedal triangle is given by = 2? cos A cos B cos C.

Centroid of a Triangle:

The point where all the three medians meet is called the centroid of a triangle.



  • Circum-center of the pedal triangle of a given triangle bisects the line joining the circum-center of the triangle to the orthocenter.

  • Orthocenter of a triangle is the same as the incentre of the pedal triangle in the same triangle.

  • If I1, I2 and I3 be the centers of the escribed circles which are opposite to A, B and C respectively and I the center of the incircle, then triangle ABC is the pedal triangle of the triangle I1I2I3 and I is the orthocenter of the triangle I1I2I3.

  • The centroid of the triangle lies on the line joining the circumventer to the orthocenter and divides it in the ratio 1: 2.


Some illustrations based on the above trigonometric concepts:


In a right-angled triangle, prove that r + 2R = s.


In a right-angled triangle, the circum-center lies on the hypotenuse.

⇒ R = a/2, where A = 90o.

Also r = (s – a) tan A/2 

= (s – a) tan45o 

= (s – a)

= s – 2R

⇒ r + 2R = s.



The ex-radii r1, r2, r3 of ΔABC are in H.P. Show that its sides a, b, c are in A.P.


The ex-radii r1, r2, r3 of ΔABC are in H.P.

Now, r1, r2, r3 are in H.P. ⇒ 2/r2 = 1/r1 + 1/r3

⇒2 (s - b)/Δ = (s – a)/Δ + (s – c)/Δ where is the area of ΔABC

⇒ 2s – 2b = 2s – a – c

⇒ 2b = a + c. Hence a, b, c are in A.P. 



Given, Δ = a2 – (b – c) 2, calculate the value of tan A.


Given, Δ = a2 – (b – c) 2

= a2 – b2 – c2 + 2bc

= 2bc – (b2 + c2 – a2)

⇒ 1/2 bc sin A = 2bc – 2bc cos A

⇒ sin A = 4 (1 – cos A)

⇒ 2 sin A/2 cos A/2 = 4 (2 sin2 A/2)

⇒ sin A/2  = 0 (not possible) or tan A/2 = 1/4 .

Now tan A = (2 tan A/2)/(1 - tan2 A/2)

                   = (1/2)/ (1 - 1/16)

                   = 8/15



Prove that the distance between the circumcenter and the orthocenter of a triangle ABC is R√1-8 cos A cos B cos C.


Let O and P be the circum-center and the orthocenter respectively.

If OF is the perpendicular to AB, we have

∠OAF = 90o – ∠AOF = 90o – C.

Also ∠PAL = 90o – C.

Hence, ∠OAP = A – ∠OAF – ∠PAL 

= A – 2(90o – C) = A + 2C – 180o

 = A + 2C – (A + B + C) = C – B.

Also OA = R, and PA = 2R cos A.

Also OA = R, and PA = 2R cos A.

Now in DAOP, 

OP2 = OA2 + PA2 – 2OA. PA cos OAP

= R2 + 4R2 cos2A – 4R2cosA cos(C–B)

= R2 + 4R2 cosA[cosA – cos(C – B)]

= R2 – 4R2 cosA[cos(B + C) + cos(C – B)]

= R2 – 8R2 cosA cosB cosC.

Hence OP = R√1 - 8 cos A cos B cos C.

You might like to refer some of the related resources listed below:

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