Question

A triangle has corners at `(2 ,3 )`, `(1 ,9 )`, and `(6 ,8 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`(x-3.0195)²+(y-6.9143)²=1.6491^2`

Explanation:

The triangle vertices `p_1=(2,3),p_2=(1,9),p_3=(6,8)`
First step is bissectrice building.
`b_1->p_1+v_1 lambda_1`
`b_2->p_2+v_2 lambda_2`

where:
`v_1 = (p_2-p_1)/norm(p_2-p_1)+(p_3-p_1)/norm(p_3-p_1) = (0.460296, 1.76726)`
`v_2 = (p_1-p_2)/norm(p_1-p_2)+(p_3-p_2)/norm(p_3-p_2)=(1.14498, -1.18251)`

The bissectrices intersection point is the circumference center calculated as the solution `(lambda_1^0,lambda_2^0)` to the system.
`p_1+v_1 lambda_1 = p_2+v_2 lambda_2`
The circumference center is obtained as
`c=p_1+v_1 lambda_1^0= p_2+v_2 lambda_2^0 = (3.01951, 6.9143)`
The circumference radius is obtained using Pythagoras.
`r = sqrt(norm(c-p_1)^2-norm((c-p_1). (p_2-p_1)/norm(p_2-p_1))^2) = 1.64914`