Radius of a Circumcircle
Given a triangle with vertices \(A\), \(B\), and \(C\), we can find a formula for the radius \(R\) of its circumcircle (circumradius).
Let \(L_1\) be the perpendicular bisector of \(BC\), \(L_2\) the perpendicular bisector of \(AC\), and \(L_3\) the perpendicular bisector of \(AB\). These three lines are concurrent at point \(O\), called the circumcenter. This point is the center of the circumcircle that passes through all three vertices \(A\), \(B\), and \(C\).
The formula of the circumradius \( R \) (radius of the circumcircle) is given by:
\[ 2R = \dfrac{BC}{\sin(A)} = \dfrac{AC}{\sin(B)} = \dfrac{AB}{\sin(C)} \]Practice Problems on Circumcircles
Problem 1: Two Methods to Find the Circumradius
A triangle has side lengths \(a = BC = 13 \, \text{cm}\), \(b = AC = 14 \, \text{cm}\) and \(c = AB = 15 \, \text{cm}\). Find the circumradius \(R\) using two different approaches.
Method 1: Heron's Formula and Area Method
- Semiperimeter: \(s = \dfrac{a+b+c}{2} = \dfrac{13+14+15}{2} = 21 \, \text{cm}\).
- Area using Heron's formula: \(\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 \, \text{cm}^2\).
- Find \(\sin(A)\) using \(\Delta = \frac{1}{2}bc\sin(A)\): \(\sin(A) = \dfrac{2\Delta}{bc} = \dfrac{2 \times 84}{14 \times 15} = \dfrac{168}{210} = \dfrac{4}{5} = 0.8\).
- Calculate \(R\): \(R = \dfrac{a}{2\sin(A)} = \dfrac{13}{2 \times 0.8} = \dfrac{13}{1.6} = 8.125 \, \text{cm}\).
Method 2: Cosine Rule and Trigonometric Identity
- Apply the cosine rule to find \(\cos(A)\): \(\cos(A) = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{14^2 + 15^2 - 13^2}{2 \times 14 \times 15}\).
- \(\cos(A) = \dfrac{196 + 225 - 169}{420} = \dfrac{252}{420} = \dfrac{3}{5} = 0.6\).
- Find \(\sin(A)\) using \(\sin^2(A) + \cos^2(A) = 1\): \(\sin(A) = \sqrt{1 - (0.6)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8\).
- Calculate \(R\): \(R = \dfrac{a}{2\sin(A)} = \dfrac{13}{2 \times 0.8} = 8.125 \, \text{cm}\).
Conclusion: Both methods yield the same circumradius \(R = 8.125 \, \text{cm}\), confirming the consistency of the formulas.
Problem 2: Right Triangle, Circumcenter and Thales' Theorem
A triangle has vertices at \(A(0,0)\), \(B(6,0)\) and \(C(0,8)\).
Part a: Find the circumcenter \(O\) and the circumradius \(R\). Verify that \(OA = OB = OC\).
Part b: Show that \(BC\) is the diameter of the circumcircle and relate this to Thales' theorem.
Solution: Part a & Part b
Part a:
- The triangle is right-angled at \(A\) because \(AB\) lies on the x-axis and \(AC\) lies on the y-axis.
- In a right triangle, the circumcenter is the midpoint of the hypotenuse \(BC\).
- \(B(6,0)\) and \(C(0,8)\): midpoint \(O = \left( \dfrac{6+0}{2}, \dfrac{0+8}{2} \right) = (3, 4)\).
- Circumradius \(R\) is the distance from \(O\) to any vertex:
\(R = OA = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5\).
\(OB = \sqrt{(3-6)^2 + (4-0)^2} = \sqrt{9 + 16} = 5\).
\(OC = \sqrt{(3-0)^2 + (4-8)^2} = \sqrt{9 + 16} = 5\). - Thus \(OA = OB = OC = R = 5\), confirming \(O\) is equidistant from all three vertices.
Part b:
- Length of hypotenuse \(BC = \sqrt{(6-0)^2 + (0-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\).
- Diameter of the circumcircle \(= 2R = 2 \times 5 = 10\).
- Since \(BC = 2R\), side \(BC\) is exactly the diameter of the circumcircle.
- Thales' theorem states that an angle inscribed in a semicircle is a right angle. Here, \(\angle BAC = 90^\circ\) and it subtends the diameter \(BC\). Therefore, points \(A\), \(B\) and \(C\) lie on a circle with diameter \(BC\), proving Thales' theorem for this triangle.