0
$\begingroup$

Given $A=(a_1,a_2), B=(b_1,b_2), C=(c_1,c_2)$, is there a simple formula to express the radius of the circumcircle of $ABC$?

Note that you could compute the radius from the sidelengths as $\frac{abc}{\sqrt{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)}}$, but I'm really hoping that there's something simpler than that. If it helps, assume $C=(0,0)$.

$\endgroup$
  • $\begingroup$ Are you looking for a formula in terms of $a,b,c$ or using coordinate geometry? $\endgroup$ – Dr. Mathva Apr 6 at 21:42
0
$\begingroup$

The square root in the denominater is equal to the twice the area of $\triangle ABC$ (Heron's formula). The same area can be calculated by embeding the plane in 3-dimensional space and using the vector product: $$ S = \frac12 |\vec{CA} \times \vec{CB}| $$ Using Cartesian coordinates, we have $$ |\vec{CA} \times \vec{CB}| = |(a_1-c_1)(b_2-c_2)-(a_2-c_2)(b_1-c_1)| = \\ = |a_1(b_2-c_2)+b_1(c_2-a_2)+c_1(a_2-b_2)| $$ so $$ R = \frac{\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}{|a_1(b_2-c_2)+b_1(c_2-a_2)+c_1(a_2-b_2)|}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.