Trigonometry

Radii and Properties of Circumcircle, Incircle and Excircle

This section covers fundamental geometric relationships, trigonometric formulas, and radius expressions for circumcircles, incircles, and excircles of triangles. This topic.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Fundamental Radius Relations of Triangle Circles

For any triangle $ABC$ with sides $a, b, c$, semi-perimeter $s = \frac{a+b+c}{2}$, area $\Delta$, circumradius $R$, inradius $r$, and excircle radii $r_1, r_2, r_3$, the fundamental identities are given by $R = \frac{abc}{4\Delta}$, $r = \frac{\Delta}{s}$, and $r_1 = \frac{\Delta}{s-a}$, with analogous formulas for $r_2$ and $r_3$.

📐 Geometric Interpretation

The circumradius $R$ measures the distance from the circumcenter (intersection of perpendicular bisectors) to any vertex. The inradius $r$ measures the perpendicular distance from the incenter (intersection of angle bisectors) to any side, acting as the radius of the inscribed circle tangent to all three sides. The exradii $r_1, r_2, r_3$ correspond to circles tangent to one side externally and the extensions of the other two sides.

⚙️ Algebraic & Structural Meaning

These relations connect symmetric algebraic polynomials of side lengths with trigonometric functions of half-angles (e.g., $r = 4R \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$), enabling swift reduction of complex trigonometric expressions in triangle geometry.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-degeneracy of Triangle	$a+b > c, \; \|a-b\| < c$	Strictly Mandatory / Necessary	If triangle inequality fails, area $\Delta = 0$, leading to division by zero in radius formulas ($R, r, r_i$ become undefined or infinite).
Angle Domain Bounds	`$A, B, C \in (0, \pi)$ with $A+B+C = \pi$`	Strictly Mandatory	Negative or obtuse values violating interior angle sum make half-angle trigonometric substitutions invalid.
Excircle Side Exclusion	$s > a, s > b, s > c$	Essential for real exradii	If $s \le a$, the excircle radius $r_1 = \frac{\Delta}{s-a}$ becomes zero or negative, violating the geometric existence of a real tangent excircle.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Product of Exradii & Inradius

$$r \cdot r_1 \cdot r_2 \cdot r_3 = \Delta^2$$

Rule: The product of the inradius and all three exradii equals the square of the area of the triangle.

Application: Use this to find $\Delta$ or ratio values instantly.

Sum of Exradii

$$r_1 + r_2 + r_3 - r = 4R$$

Rule: The sum of the three exradii minus the inradius is identically equal to four times the circumradius.

Reciprocal Sum of Exradii

$$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$$

Rule: The sum of the reciprocals of the exradii equals the reciprocal of the inradius.

Sum of Squares of Radii

$$r_1^2 + r_2^2 + r_3^2 + r^2 = 16R^2 - (a^2+b^2+c^2)$$

Rule: Relates the sum of squares of all four circle radii directly to $R$ and side length squares.

Inradius in Terms of Circumradius and Angles

$$r = 4R \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)$$

Rule: Expresses inradius purely through circumradius and half-angle sines.

Application: Used to evaluate maximum/minimum bounds in triangle inequalities.

Distance between Circumcenter and Incenter ($OI$)

$$OI^2 = R^2 - 2Rr$$

Rule: Euler's formula for the distance between circumcenter $O$ and incenter $I$. Since $OI^2 \ge 0$, it implies Euler's famous inequality $R \ge 2r$.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Inradius with Exradius Denominators

Common Pitfall: Using $s$ instead of $(s-a)$ in the denominator when calculating excircle radius $r_1$.

Counterexample: $r_1 = \frac{\Delta}{s}$ (Wrong) vs $r_1 = \frac{\Delta}{s-a}$ (Correct)

Why it fails: Leads to completely incorrect numerical calculations and mismatched options.

🛡️ Exam Defense: Remember: Inradius uses $s$ ($r = \Delta/s$), while excircle radius $r_1$ uses $(s-a)$, $r_2$ uses $(s-b)$, and $r_3$ uses $(s-c)$.

⚠️ COMMON ERROR #02

Misinterpreting Euler's Inequality Direction

Common Pitfall: Assuming $R \le 2r$ instead of $R \ge 2r$.

Counterexample: For an equilateral triangle, $R = 2r$ ($R/r = 2$), whereas for general triangles $R \ge 2r$.

Why it fails: Results in choosing wrong inequality bounds in range-based questions.

🛡️ Exam Defense: The circumradius is always at least twice the inradius ($R \ge 2r$), equality holding exclusively for equilateral triangles.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Circumradius ($R$)	Any triangle $ABC$	$R = \frac{abc}{4\Delta} = \frac{a}{2\sin A}$	Radius of circle passing through all 3 vertices	Keywords: Circumcircle, vertices, $abc/4\Delta$
Inradius ($r$)	Any triangle $ABC$	$r = \frac{\Delta}{s} = (s-a)\tan\frac{A}{2}$	Radius of circle inscribed tangent to 3 sides	Keywords: Incircle, inscribed, incenter
Exradius ($r_1$)	Any triangle $ABC$	$r_1 = \frac{\Delta}{s-a} = s \tan\frac{A}{2}$	Radius of excircle tangent to side $a$ externally	Keywords: Excircle, escribed, side $a$
Euler Distance ($OI$)	Distance between $O$ and $I$	$OI^2 = R^2 - 2Rr \implies R \ge 2r$	Distance squared between circumcenter and incenter	Keywords: Distance between centers, $R \ge 2r$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

In a triangle $ABC$, if $a=3, b=4, c=5$, find the circumradius $R$ and inradius $r$.

(A) $R = 2.5, r = 1$(B) $R = 2, r = 1.5$(C) $R = 5, r = 1$(D) $R = 2.5, r = 2$

📚 Method A: Formal Textbook Derivation

Step 1: Check triangle type ($3^2+4^2=5^2$, right-angled triangle). Step 2: Area $\Delta = \frac{1}{2} \times 3 \times 4 = 6$. Step 3: Semi-perimeter $s = \frac{3+4+5}{2} = 6$. Step 4: $R = \frac{abc}{4\Delta} = \frac{3 \times 4 \times 5}{4 \times 6} = \frac{60}{24} = 2.5$. Step 5: $r = \frac{\Delta}{s} = \frac{6}{6} = 1$.

⚡ Method B: 25-Second Exam Shortcut

For a right-angled triangle, circumradius $R$ is always half of the hypotenuse ($5/2 = 2.5$), and inradius $r = \frac{b+c-a}{2} = \frac{3+4-5}{2} = 1$.

💡 Tip: Recognize Pythagorean triplets instantly to bypass lengthy Heron's formula calculations.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If in a triangle $ABC$, $r_1 = 2, r_2 = 3, r_3 = 6$, find the inradius $r$.

(A) $1$(B) $1.5$(C) $2$(D) $0.5$

📚 Method A: Formal Textbook Derivation

Step 1: Use the reciprocal relation $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$. Step 2: Substitute values: $\frac{1}{r} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$. Step 3: Simplify: $\frac{1}{r} = \frac{3+2+1}{6} = \frac{6}{6} = 1$. Step 4: Therefore, $r = 1$.

⚡ Method B: 25-Second Exam Shortcut

Direct application of harmonic sum of exradii reciprocals equals inradius reciprocal.

💡 Tip: Memorize standard harmonic formulas for exradii to solve.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: In any triangle $ABC$, the value of $r_1 r_2 + r_2 r_3 + r_3 r_1$ is equal to:

A $s^2$

B $4Rr + r^2$

C $s(r_1+r_2+r_3)$

D $bc + ca + ab$

Derivation & Analysis:

Using standard trigonometric identities for exradii, $r_1 r_2 + r_2 r_3 + r_3 r_1 = s^2$.

MCQ 2: If the sides of a triangle are $13, 14, 15$, the inradius $r$ is:

A $3$

B $4$

C $5$

D $6$

Derivation & Analysis:

Semi-perimeter $s = \frac{13+14+15}{2} = 21$. Area $\Delta = \sqrt{21(8)(7)(6)} = \sqrt{21 \times 336} = 84$. Inradius $r = \frac{\Delta}{s} = \frac{84}{21} = 4$.

MCQ 3: For an equilateral triangle of side $a$, the ratio of circumradius $R$, inradius $r$, and exradius $r_1$ is:

A $1 : 2 : 3$

B $2 : 1 : 3$

C $3 : 1 : 2$

D $1 : 1 : 1$

Derivation & Analysis:

For an equilateral triangle, $R = \frac{a}{\sqrt{3}}$, $r = \frac{a}{2\sqrt{3}}$, and $r_1 = \frac{\sqrt{3}a}{2}$. Thus $R : r : r_1 = \frac{1}{\sqrt{3}} : \frac{1}{2\sqrt{3}} : \frac{\sqrt{3}}{2} = 2 : 1 : 3$.

MCQ 4: The expression $4R \sin A \sin B \sin C$ is identically equal to:

A $s$

B $r$

C $\frac{\Delta}{R}$

D $a \sin B$

Derivation & Analysis:

We know $a = 2R \sin A$, etc. Also $r = 4R \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$, but $4R \sin A \sin B \sin C = \frac{abc}{R} = \frac{4\Delta}{R}$ is related to area, while $4R \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = r$. Wait, note that $4R \prod \sin A = \frac{abc}{2R^2}$... Actually, using projection formulas, $4R \sin A \sin B \sin C$ simplifies to $\frac{abc}{R^2}$ or via standard results equals $r(\cot\frac{A}{2}...)$ - correct standard identity is $r = 4R \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.

MCQ 5: If $r_1, r_2, r_3$ are exradii of triangle $ABC$ opposite to sides $a, b, c$, then $\sum r_2 r_3$ equals:

A $s^2$

B $4Rr + r^2 + s^2$

C $s(r_1+r_2+r_3)$

D $s^2 + 4Rr + r^2$

Derivation & Analysis:

Standard trigonometric identity in triangle properties states that $r_1 r_2 + r_2 r_3 + r_3 r_1 = s^2$.

MCQ 6: In a triangle $ABC$, if $a=2, b=3, c=4$, what is the value of $\Delta$?

A $\frac{3\sqrt{15}}{4}$

B $\frac{15}{4}$

C $\sqrt{15}$

D $\frac{\sqrt{15}}{2}$

Derivation & Analysis:

$s = \frac{2+3+4}{2} = \frac{9}{2}$. $\Delta = \sqrt{\frac{9}{2}(\frac{5}{2})(\frac{3}{2})(\frac{1}{2})} = \sqrt{\frac{135}{16}} = \frac{3\sqrt{15}}{4}$.

MCQ 7: Euler's relation for the distance $OI$ between incenter and circumcenter is:

A $OI^2 = R^2 + 2Rr$

B $OI^2 = R^2 - 2Rr$

C $OI^2 = 2R^2 - r^2$

D $OI = R - 2r$

Derivation & Analysis:

Euler's theorem states $OI^2 = R^2 - 2Rr$. Since $OI^2 \ge 0$, this yields $R \ge 2r$.

MCQ 8: If the inradius $r = 2$ and circumradius $R = 5$, the distance between incenter and circumcenter is:

A $3$

B $4$

C $\sqrt{5}$

D $2$

Derivation & Analysis:

$OI^2 = R^2 - 2Rr = 5^2 - 2(5)(2) = 25 - 20 = 5$... wait, let's adjust numbers: if $R=5, r=2$, $OI^2 = 25 - 20 = 5$ (not a square). Let's use $R=5, r=4$, then $OI^2 = 25 - 40$ (invalid since $R \ge 2r \implies 5 \ge 8$ false). Let's take $R=5, r=2$, distance squared is $5$, wait option check: if $R=5, r=2$, $OI = \sqrt{5}$. Let's use standard exam numbers: $R=5, r=2$ gives $OI = \sqrt{5}$.

MCQ 9: The ratio of the area of the circumcircle to the area of the incircle of an equilateral triangle is:

A $2 : 1$

B $4 : 1$

C $8 : 1$

D $16 : 1$

Derivation & Analysis:

For an equilateral triangle, $R = 2r$. Area ratio = $\frac{\pi R^2}{\pi r^2} = \left(\frac{R}{r}\right)^2 = 2^2 = 4 : 1$.

Q10

MCQ 10: In any triangle $ABC$, $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$ is equal to:

A $\frac{1}{r}$

B $\frac{1}{R}$

C $\frac{2}{r}$

D $\frac{s}{\Delta}$

Derivation & Analysis:

By definition, $\sum \frac{1}{r_1} = \sum \frac{s-a}{\Delta} = \frac{3s - (a+b+c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.

Common questions

Geometrically, an excircle touches one side of the triangle externally while the other two sides are produced. The tangents drawn from the opposite vertex to the excircle have length equal to the semi-perimeter $s$, making the effective distance components $(s-a)$ for the corresponding side.

Calculation errors in Heron's formula and mixing up the formulas for circumradius ($abc/4\Delta$) versus inradius ($\Delta/s$). Memorizing the half-angle and sine rule linkages prevents these slips.

Master standard triangle types (equilateral, right-angled, isosceles) and memorize key ratio constants like $R:r:r_1 = 2:1:3$ for equilateral triangles and Euler's inequality $R \ge 2r$.

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