Circles connected with a given Triangle

The circle which passes through three vertices of a triangle is called the circumscribing circle or circumcircle.

The center of this circle is called circumcenter, usually denoted as O, and its radius is called circumradius, usually denoted as R. It can be shown that circumcenter is the point of intersection of right bisectors of the sides of a triangle.

The (unique) circle which is drawn within a triangle so as the three sides touch this circle is called the inscribed circle or incircle. The center of this circle is called incenter, usually denoted by I, and its radius is called inradius, usually denoted by r. It can be shown that incenter is the point of concurrence of the bisectors of the three (internal) angles of the triangle.

The circle which lies outside the triangle and touches the side BC and also the sides AB and AC produced is called escribed circle or excircle opposite to angle A. Its center is called excenter and its radius is called exradius. It can be shown that excenter is the point of concurrence of internal bisector of angle A and external bisectors of angles B and C. Similarly there are two more excircles, one opposite to angle B and one opposite to angle C. The three excenters are usually denoted as I₁, I₂, I₃ and the three ex-radii are usually denoted as r₁, r₂, r₃.
        
       

The Circumradius

In any ABC,

1. R = a/(2 sin A) = b/(2 sin B) = c/(2 sin C), and
2. R = abc/4

The Inradius

In any ABC,

1. r = /s
2. r = (s -a) tan A/2 = (s -b) tan B/2 = (s -c) tan C/2
3.  

The Exradius

In any ABC,

1. r₁ = /(s-a) = s tan A/2 = (s -c) cot B/2 = (s -b) cot C/2
2.  

Illustrative Examples

Example

In any ABC, prove that
    1/r₁ +1/r₂ +1/r₃ = 1/r

Solution

 
= (3s - 2s)/ = s/  = 1/r

Example

In a ABC, a = 18, b = 24, c = 30 cms. Find:

1. 
2. R
3. r
4. r₁, r₂ , r₃
5. Area of circumcircle
6. Area of incircle
7. Area of escribed circle opposite angle B

Solution

2s = a +b +c = 18 +24 +30 = 72 cm =>    s = 36 cm

1. = [s (s -a)(s -b)(s -c)] = [36.18.12.6] = 216 cm²
2. R = abc/4 = (18.24.30)/(4.216) = 15 cm
3. r = /s = 216/36 = 6 cm
4. r₁ = /(s-a) = 216/18 = 12 cm,
   r₂ = /(s-b) = 216/12 = 18 cm,
   r₃ = /(s-c) = 216/6 = 36 cm
5. Area of circumcircle = R² = (15)² = 225 cm²
6. Area of incircle = r² = (6)² = 36 cm²
7. Area of escribed circle opposite to angle B
              = r₂² = (18)² = 344 cm²

Exercise

In a ABC, prove the following (1 -9):

1. s = 4 R cos A/2 cos B/2 cos C/2
2. = 2R² sin A sin B sin C
3. (a +b) sec [(A-B)/2] = 4R cos C/2
4. R(a² +b² +c²) = abc(cot A +cot B +cot C)
5. (s/a -1)(s/b -1)(s/c -1) = r/R
6. (b +c) tan A/2 +(c +a) tan B/2 + (a +b) tan C/2 = 4(R +r)
7. 8Rr(cos²A/2 +cos²B/2 +cos²C/2) = 2(ab +bc +ca)-(a² +b² +c²)
8. (i) r r₁ r₂ r₃ = ²
   (ii) (r₁ -r) (r₂ -r) (r₃ -r) = 4 R² r
   (iii) rr₁/(r₂r₃) = tan² A/2
9. (r₁ +r₂)(r₂ +r₃)(r₃ + r₁) = 4 R s²
10. If r₁ = r₂ +r₃ +r, prove that A = 90°
11. If a² +b² +c² = 8 R², prove that the triangle is right angled.
12. If the diameter of an excircle is equal to the perimeter of the triangle, show that the triangle is right angled.
13. If in a ABC, a = 4 cm, b = 5 cm, c = 6 cm, calculate
    (i)      (ii) R      (iii) r       (iv) r₁, r₂, r₃
    (v) lengths of altitudes p₁, p₂, p₃
    (vi) area of circumcircle (correct upto 2 decimals. Use = 22/7)
    (vii) area of incircle
    (viii) area of escribed circle opposite to angle C
    (ix) length of bisector of angle A
    (x) sin A             (xi) sin A/2
14. If in a ABC, a : b : c = 5 : 6 : 7, find the ratio of the radius of the circumcircle to that of incircle.
15. If the area of a triangle is 24 sq. cm. and its ex-radii are 4 cm, 1 cm, 12 cm, find the lengths of its three sides.
16. If the angles of a triangle are in ratio 1 : 2 : 3 and its circumradius is 10 cm, show that the lengths of its sides are 10 cm, 20 cm, 103 cm.

Answers