Tenth Grade Mathematics: Inscribed and Circumscribed Circles

Every triangle in Euclidean geometry defines two unique circles: the circumscribed circle (circumcircle), which passes through all three of its vertices, and the inscribed circle (incircle), which lies entirely within the triangle and touches all three of its sides. Constructing these circles requires locating the concurrency points of the triangle's perpendicular bisectors and internal angle bisectors respectively. In this chapter, we master the compass-and-straightedge techniques for constructing circumcircles and incircles, establishing the algebraic formulas for their radii and centers.

Construction of the Circumscribed Circle (Circumcircle)

The circumscribed circle (or circumcircle) of a triangle is the unique circle that passes through all three vertices A, B, and C of the triangle.

The Geometric Center Principle:
The center of the circumcircle (the Circumcenter, denoted by O) must be
equidistant from all three vertices:
                           OA = OB = OC = R
where R is the circumradius.

Recall that the locus of all points equidistant from two points A and B is the perpendicular bisector of segment AB. Therefore, the circumcenter O is the point of concurrency of the perpendicular bisectors of the sides of the triangle.

Step-by-Step Circumcircle Construction:
Step 1: Construct the given triangle ABC according to specified dimensions.
Step 2: Construct the perpendicular bisector of side AB using compass arcs.
Step 3: Construct the perpendicular bisector of side BC using compass arcs.
Step 4: The intersection point of these two perpendicular bisectors is the
        Circumcenter O. (The third perpendicular bisector automatically passes
        through O).
Step 5: Place the compass point at O, adjust the compass pencil to any vertex
        (say OA), and draw the complete circle.
Conclusion: The circle passes through A, B, and C, forming the circumcircle.

Position of the Circumcenter O:
- Acute-angled Triangle: Circumcenter O lies STRICTLY INSIDE the triangle.
- Right-angled Triangle: Circumcenter O lies at the MIDPOINT of the hypotenuse
  (Circumradius R = Hypotenuse / 2).
- Obtuse-angled Triangle: Circumcenter O lies STRICTLY OUTSIDE the triangle.

Construction of the Inscribed Circle (Incircle)

The inscribed circle (or incircle) of a triangle is the circle that lies completely inside the triangle and touches all three sides AB, BC, and CA tangentially.

The Geometric Center Principle:
The center of the incircle (the Incenter, denoted by I) must be equidistant
from all three sides of the triangle:
                   Distance to AB = Distance to BC = Distance to CA = r
where r is the inradius.

Recall that the locus of all points equidistant from two intersecting lines is the internal bisector of the angle between them. Therefore, the incenter I is the point of concurrency of the internal angle bisectors of the triangle.

Step-by-Step Incircle Construction:
Step 1: Construct the given triangle ABC.
Step 2: Construct the internal angle bisector of Angle B using compass arcs.
Step 3: Construct the internal angle bisector of Angle C using compass arcs.
Step 4: The intersection point of these two angle bisectors is the Incenter I.
Step 5: Drop a perpendicular from I to any side (say side BC) to locate the point
        of contact D. The length ID is the exact inradius r!
Step 6: Place the compass point at I, set radius to ID, and draw the circle.
Conclusion: The circle touches sides AB, BC, and CA tangentially, forming the incircle.

ASCII Incircle Schema:
                         A
                        /                        / .                       /  I  \    I = Intersection of Angle Bisectors
                     /  /|\  \   ID = Inradius r (Perpendicular to BC)
                  B +----+----+ C
                         D

Algebraic Formulas for Inradius and Circumradius

For a triangle with side lengths a, b, c, semi-perimeter s = (a + b + c)/2, and area Delta:

1. Inradius Formula:
                          r = Delta / s

2. Circumradius Formula:
                             a * b * c
                        R = -----------
                             4 * Delta

3. For a Right-Angled Triangle (legs a and b, hypotenuse c):
                          a + b - c
                     r = -----------     and     R = c / 2
                              2

Chapter Exercises

Exercise 1: Construct the circumcircle of an equilateral triangle of side 6 cm. Measure the circumradius and verify its length using the formula R = a / sqrt(3).

Exercise 2: Construct the incircle of a right-angled triangle ABC whose legs containing the right angle are AB = 6 cm and BC = 8 cm. Measure the inradius and verify algebraically.

Exercise 3: Prove algebraically that for any right-angled triangle with legs a and b and hypotenuse c, the inradius is given by r = (a + b - c) / 2.

Solutions and Explanations

Solution 1: Draw an equilateral triangle ABC with each side equal to 6 cm. Construct the perpendicular bisectors of sides BC and AC. Their intersection point is the circumcenter O. With O as center and radius OA, draw the circumcircle. The circle passes through A, B, and C. Measuring the circumradius OA gives R = 3.46 cm. Algebraic verification: for an equilateral triangle, area Delta = (sqrt(3)/4)a^2 = (sqrt(3)/4)(36) = 9sqrt(3) cm^2. Using the circumradius formula: R = (a * b * c) / (4 * Delta) = (6 * 6 * 6) / (4 * 9sqrt(3)) = 216 / (36sqrt(3)) = 6 / sqrt(3) = 2sqrt(3) cm. Since sqrt(3) is approximately 1.732, R = 2 * 1.732 = 3.464 cm. The measured radius matches the formula exactly.

Solution 2: Construct right-angled triangle ABC with base BC = 8 cm, Angle B = 90 degrees, and vertical leg AB = 6 cm. Hypotenuse AC = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10 cm. Construct the angle bisectors of Angle B and Angle C. Their intersection point is the incenter I. From I, drop a perpendicular to BC meeting BC at D. Measure ID = r; the measured inradius is 2 cm. With I as center and radius ID = 2 cm, draw the incircle. Algebraic verification: Area Delta = (1/2) * 8 * 6 = 24 cm^2. Semi-perimeter s = (6 + 8 + 10)/2 = 24/2 = 12 cm. Inradius r = Delta / s = 24 / 12 = 2 cm. The algebraic value verifies the construction.

Solution 3: Let the right triangle have legs a and b and hypotenuse c, with the right angle at vertex C. The incircle with center I and radius r touches side BC at P, side AC at Q, and hypotenuse AB at R. Radii IP and IQ are perpendicular to sides BC and AC. In quadrilateral IPCQ, Angle C = 90 degrees, Angle IPC = 90 degrees, and Angle IQC = 90 degrees, and IP = IQ = r. Therefore, IPCQ is a square of side length r. This implies CP = r and CQ = r. By the equal tangent lengths theorem: the tangents from B are equal: BP = BR = a - r. The tangents from A are equal: AQ = AR = b - r. The hypotenuse c is the sum of AR and BR: c = AR + BR = (b - r) + (a - r) = a + b - 2r. Rearranging this equation to solve for r: 2r = a + b - c ==> r = (a + b - c) / 2. This proves the formula.