The equation of the circle circumscribing the triangle formed by the lines $y = 0, y = x$ and $2x + 3y = 10$ is?
I know this can be done by solving two equations at a time and finding the vertex. Then forming 3 different equations to solve for the centre (abscissa, ordinate) and the radius( which are 3 variables).
I would like to know if there's a shorter method to find the equation of the circle for any 3 random lines given. (It's easier in this particular case as one of the coordinates is $(0,0)$ which makes $c=0$)
Best Answer
The equation of the circle is the following:
$$\operatorname{det}\left( \begin{bmatrix} \dfrac{1}{y} & 0 & 1 \\ \dfrac{2}{x-y} & 1 & -1 \\ \dfrac{13}{2x+3y-10} & 2 & 3\end{bmatrix} \right) = 0 \Longrightarrow \left(x-\dfrac{5}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2 = \dfrac{13}{2}$$
This comes from the generic formula that if you have three lines:
$$a_rx+b_ry+c_r=0, r=1,2,3$$
whose area of intersection forms a triangle, then the equation of the circle circumscribing the triangle is given by:
$$\operatorname{det}\left(\begin{bmatrix} \dfrac{a_1^2+b_1^2}{a_1x+b_1y+c_1} & a_1 & b_1 \\ \dfrac{a_2^2+b_2^2}{a_2x+b_2y+c_2} & a_2 & b_2 \\ \dfrac{a_3^2+b_3^2}{a_3x+b_3y+c_3} & a_3 & b_3\end{bmatrix}\right) = 0$$
Check out this post here for a proof.