In-Circle and In-Centre: Formula, Properties and Examples

A triangle is more special as compared to other polygons as it is the polygon having the least number of sides. A triangle has six main elements, three sides, and three angles. There are many properties of triangles like circumcentre, incentre, centroid, orthocentre, etc. The incircle is the largest inscribed circle of a triangle. In real life, we use encircles in the design of gears.

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In this article, we will cover the concept of the Circumcircle. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of five questions have been asked on this concept including one in 2021.

What is In-Circle and In-Centre?

The incircle or circle inscribed in a triangle is the largest circle contained in the triangle; it touches (or it is tangent to) the three sides of the triangle. The center / of the incircle is called the incentre of the triangle. The radius of the incircle is called inradius and is usually denoted by r .

The point of intersection of the internal angle bisectors of a triangle is called the in-center of the triangle. Also, a circle that can be inscribed within a triangle such that it touches each side of the triangle internally is called an incircle triangle. Its center is the center of the given triangle. The in-center of a triangle is denoted by l .

In Radius Formula

The radius of the in-circle of a triangle is called the in-radius and it is denoted by ‘r’.

1. In - radius, $r$ is given by $=\frac{\mathrm{\Delta }}{s}$
2. $r=(s-a)\tan \frac{\mathrm{A}}{2}=(s-b)\tan \frac{\mathrm{B}}{2}=(s-c)\tan \frac{\mathrm{C}}{2}$
3. $r=4\mathrm{R}\sin \frac{\mathrm{A}}{2}\sin \frac{\mathrm{B}}{2}\sin \frac{\mathrm{C}}{2}$

Derivation of In Radius

1. Consider the triangle, $ABC$

We know that, the area of $ABC=$ area of $IBC+$ area of $IBA+$ area of $ICA$

$\begin{array}{rl}\mathrm{\Delta }& =\frac{1}{2}ar+\frac{1}{2}br+\frac{1}{2}cr\\ & =\frac{1}{2}r(a+b+c)\\ & =\frac{1}{2}r(2s)[\because 2s=a+b+c]\\ & =rs\end{array}$

2. From the half-angle formula of tangent

$\tan \frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{s}(\mathrm{s}-\mathrm{a})}}$

Multiply both sides with ( $s$ - a)

$\begin{array}{rl}(s-a)\tan \frac{\mathrm{A}}{2}& =(s-a)\sqrt{\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{s}(\mathrm{s}-\mathrm{a})}}\\ & =\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\\ & =\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}\\ & =\frac{\mathrm{\Delta }}{s}=r[\because \mathrm{\Delta }=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})]}\end{array}$

In a similar fashion, other formulae can be proved

3. From the half-angle formula of sine

$\begin{array}{rl}& \sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}\\ & \sin \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{ac}}\\ & \sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}\end{array}$

So,
$[$ We are using the fact that $\mathrm{\Delta }=\frac{\mathrm{abc}}{4\mathrm{R}}$ and $\mathrm{\Delta }=\sqrt{s(s-a)(s-b)(s-c)}]$

$\begin{array}{rl}& 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}\\ & =4\mathrm{R}\frac{(s-a)(s-b)(s-c)}{abc}\\ & =\frac{4\mathrm{R}}{\mathrm{abc}}\frac{s(s-a)(s-b)(s-c)}{s}\\ & =\frac{1}{\mathrm{\Delta }}\frac{{\mathrm{\Delta }}^2}{\mathrm{s}}=\frac{\mathrm{\Delta }}{\mathrm{s}}=\mathrm{r}\end{array}$

Length of tangent from vertices to the In-circle

In the figure, $\mathrm{AF}=\mathrm{AE}=\mathrm{x}$

$\begin{array}{rl}& BF=BD=y\\ & CE=CD=z\\ & AB+BC+CA=AF+FB+BD+DC+CE+EA\\ & a+b+c=2x+2y+2z\\ & 2s=2x+2y+2z\\ & s=x+y+z\end{array}$

Now, $y+z=BC=a$. So, $x-s=a$
Similarly, $y=s-b$ and $z=s-c$
Thus, lengths of tangents to incircle from the vertice $A,B$, and $C$ are(s-a),(sb), and (s-c) respectively

Distance of incentre from the vertices of the triangle

$\mathrm{Al}=\frac{r}{\sin \frac{A}{2}}$

Similarly, $\mathrm{BI}=\frac{r}{\sin \frac{B}{2}}$ and $\mathrm{CI}=\frac{r}{\sin \frac{C}{2}}$
Length of angle Bisector

$\mathrm{AP}=\frac{2bc}{b+c}\ast \cos \frac{A}{2}$

Similarly, $\mathrm{BQ}=\frac{2ac}{a+c}\ast \cos \frac{B}{2}$ and $\mathrm{CR}=\frac{2ab}{b+a}\ast \cos \frac{C}{2}$

Summary

The incircle enhances geometric reasoning and problem-solving capabilities, providing a bridge between theoretical concepts and practical applications in various scientific and engineering disciplines. Its study not only deepens our understanding of triangle geometry but also enriches our appreciation of fundamental mathematical principles.

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Solved Examples Based on In-Circle and In-Centre

Example 1: To find the coordinates of the incenter of a triangle whose vertices are given as $A(-36,15)B(0,0)$, and $C(20,15)$.
Solution: Given that,

$\begin{array}{rl}& A(-36,15)=(x_1,y_1)\\ & B(0,0)=(x_2,y_2)\\ & C(20,15)=(x_3,y_3)\end{array}$
To find the length of the sides by using the distance formula,

$\begin{array}{rl}& AB=c=\sqrt{(-36-0{)}^2+(15-0{)}^2}\\ & AB=c=39\\ & BC=a=\sqrt{(0-20{)}^2+(0-15{)}^2}\\ & BC=a=25\\ & CA=b=\sqrt{(20+36{)}^2+(15-15{)}^2}\\ & CA=b=56\end{array}$

Substituting the values in the incenter formula,

$\begin{array}{rl}& (\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})\\ & (\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})=(\frac{-900+0+780}{120},\frac{375+0+585}{120})\\ & (\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})=(-1,8)\end{array}$

Hence, the answer is (-1, 8)

Example 2: Let the centroid of an equilateral triangle $ABC$ be at the origin. Let one of the sides of the equilateral triangle be along the straight line $x+y=3$. If $R$ and $r$ be the radius of circumcircle and incircle respectively of $\mathrm{△}ABC$ then $(R+r)$ is equal to:

Solution

$\begin{array}{rl}& \mathrm{r}=\mathrm{OM}=\frac{3}{\sqrt{2}}\\ & \sin {30}^{\circ }=\frac{1}{2}=\frac{\mathrm{r}}{\mathrm{R}}\\ & \Rightarrow \mathrm{R}=\frac{6}{\sqrt{2}}\\ & \therefore \mathrm{r}+\mathrm{R}=\frac{9}{\sqrt{2}}\end{array}$

Hence, the answer is $\frac{9}{\sqrt{2}}$
Example 3: Let $\mathrm{a},\mathrm{b}$, and c be the length of sides of a triangle ABC such that $\frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9}$
If $r$ and $R$ are the radius of the incircle and radius of the circumcircle of the triangle $ABC$, respectively, then the value of $\frac{R}{r}$ is equal to:
[JEE MAINS 2022]
Solution
Let $\frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9}=\mathrm{k}$
So $\mathrm{a}+\mathrm{b}=7\mathrm{k},\mathrm{b}+\mathrm{c}=8\mathrm{k},\mathrm{c}+\mathrm{a}=9\mathrm{k}$
To solve this we get $\mathrm{a}=4\mathrm{k},\mathrm{b}=3\mathrm{k},\mathrm{c}=5\mathrm{k}$ and calculate semi-perimeter $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\Rightarrow \mathrm{s}=6\mathrm{k}$ we know that $\frac{\mathrm{r}}{\mathrm{R}}=4\sin \frac{\mathrm{A}}{2}\cdot \sin \frac{\mathrm{B}}{2}\sin \frac{\mathrm{C}}{2}$

Hence, the answer is

$\begin{array}{rl}& \Rightarrow \frac{r}{R}=4\sqrt{\frac{(s-b)(s-c)}{bc}}\times \sqrt{\frac{(s-a)(s-c)}{ac}}\times \sqrt{\frac{(s-a)(s-b)}{ab}}\\ & \Rightarrow \frac{r}{R}=4\cdot \frac{(s-a)(s-b)(s-c)}{abc}\\ & \Rightarrow \frac{r}{R}=4\cdot \frac{(2k)(3k)(k)}{(4k)(3k)(5k)}\Rightarrow \frac{r}{R}=\frac{2}{5}\end{array}$

So the value of $\frac{R}{r}=\frac{5}{2}$
Hence, the answer is $\frac{5}{2}$