BodhiAI

Nov. 12, 2019

**Radius of Circles Connected with A Triangle**

**A. Circumscribed Circle and Its Radius **

**The circle passing through the vertices of ΔABC is called circumscribing circle or circumcircle of this triangle. The centre of this circle is called circumcentre and is usually denoted by O. It is the point where the perpendicular bisectors of the sides of the triangle meet. The radius of this circle is called circum-radius and is usually denoted by R. **

**In the figure, OB = OC = R **

**Since, OD is perpendicular to BC, we have **

** **

**Also, the angle subtended by an arc of a circle at the centre is double of the angle subtended by it at the circumference **

** **

** **

**Note that the triangle BOD and COD are congruent. **

**In right angled triangle BOD, we have**

** **

**Hence, by using SINE Rule, we have **

** ......(1)**

**Also, area of **

** **

** .......(2)**

**B. Inscribed Circle And Its Radius**

**The circle which touches all the three sides of the ΔABC internally is called the inscribed circle or the incircle of this triangle. The centre of this circle is known as incentre and is usually denoted by I. It is the point where the internal bisectors of the angles of the triangle meet. The radius of this circle is known as in-radius and is denoted by r.**

**Now, in the figure ID, IE and IF are perpendiculars to the side BC, CA and AB respectively.**

**Hence, ID = IE = IF = r {by definition}**

**Now, ar (ΔABC) = ar **

** **

** **

** ......(3)**

**Also, we know that the lengths of tangents to a circle from an external point are equal **

** BD = BF, CD = CE, AE = AF **

**Let BD = β , CE =α and AE =γ **

**Then perimeter of **

** a + b + c **

** **

**Now, BD = = **

** BD = s - **

**Similarly, CD = s - c and AE = s - a**

**Now from we have **

** **

** **

**Similarly, and **

**Thus, .....(4)**

**Again, **

** **

** **

** **

** **

**Similarly, ......(5)**

**Now, {using (5)}**

** {using (1)}**

** **

** .......(6)**

**C. Escribed Circles and Their Radii **

**The circle which touches the side BC of the ΔABC and the other two sides AB and AC produced is called the escribed circle or ex-circle opposite to the angle A. The centre of this circle is known as ex-centre and is usually denoted by I1. It is the point where the external bisectors of the angles B and C and the internal bisector of the angle A meet. The radius of this circle is known as ex-radius and is denoted by r1. Similarly the centres of escribed circles opposite to the angle B and C are denoted by I2 and I3 respectively and their radii by r2 and r3 respectively. In the figure I1D is perpendicular to BC while I1E and I1F are Since, I1D = I1E = I1F = r1**

**Now, ar **

** **

** **

** **

** ......(7)**

**Similarly, **

**Since the lengths of tangents to a circle from an external point are equal, we have **

** AE = AF, BD = BF, CD = CE**

**Now, AE + AF = (AC + CE) + (AB + BF)**

** = (AC+ CD) + (AB + BD)**

** = AC + AB + (BD + CD)**

** = AC + AB + BC **

** = perimeter of the triangle ABC = 2s**

**Since, AE = AF, we have **

** 2AE = 2AF = 2s **

**Now from we have **

** **

** **

**Similarly, ........(8)**

**Again, **

**Similarly, **

**Now, a = BC = BD + CD **

** **

** **

** **

** **

** **

**Similarly, ......(9)**

**Now, {using (9)}**

** {using (1)}**

** **

** **

**Similarly, **

**and **

**D. Orthocentre of a Triangle **

**In the figure AD, BE and CF are altitudes of the ΔABC i.e. are perpendiculars from the vertices A, B, C to the opposite sides. We know from geometry that these three perpendiculars are concurrent and their point of concurreny P is called the orthocentre of the ΔABC .The triangle DEF is called the pedal triangle of **

**The distances of the orthocentre from the vertices and the sides. **

**In right angled at E, we have **

** PA = AE sec **

** **

**But from **

** PA = c cos A cosec C**

** PA = 2R sin C cos A {using(1)}**

** PA = 2R cosA**

**Now, since PA = 2R cos B, and PC = 2R cos C**

**Again, from we have **

** PD = BD tan **

** PD = BD tan **

** PD = BD cot C = AB cos B cot C = c cos B cot C **

** PD = 2R sin C cos B. **

** PD = 2R cosB cosC**

**Similarly, PE = 2R cosC cosA, **

**and PF = 2R cosA cosB **

**E. Bisectors of the Angles **

**In the figure AD is the bisector of the angle A and divies BC into two portions of lengths x and y. Since, we know that the bisector of an angle divides the opposite side in the ratio of the adjacent sides.**

** **

** **

** **

**Again let l be the length of the bisector AD **

**We have, **

** **

** **

** **

**Similarly, **

**and **