When point is inside of circle

Question

Let we have points $A_1,A_2,A_3,A_4$.In time $t$ point $A_i$ has coordinates $(x_i,y_i) + (v_{xi},v_{yi}) * t$, all parametrs are given. Describe algorithm to find all $t$ when point $A_4$ inside circle circumscribed around $A_1,A_2,A_3$ , or to find first moment when this happens.

Answer 1

Let $A_i(X_i,Y_i)$ where $$X_i=x_i+v_{xi}t\quad\text{and}\quad Y_i=y_i+v_{yi}t$$ for $i=1,2,3$ and $4$.

According to MathWorld, the center of the circle passing through three points $(X_1,Y_1),(X_2,Y_2)$ and $(X_3,Y_3)$ is given by $$\left(-\frac{b}{2a},-\frac{c}{2a}\right)$$ and its radius is given by $$\sqrt{\frac{b^2+c^2}{4a^2}-\frac da}$$ where $$a=\begin{vmatrix} X_1 & Y_1 & 1 \\ X_2 & Y_2 & 1 \\ X_3 & Y_3 & 1 \\ \end{vmatrix},\ b=-\begin{vmatrix} X_1^2+Y_1^2 & Y_1 & 1 \\ X_2^2+Y_2^2 & Y_2 & 1 \\ X_3^2+Y_3^2 & Y_3 & 1 \\ \end{vmatrix},$$ $$c=\begin{vmatrix} X_1^2+Y_1^2 & X_1 & 1 \\ X_2^2+Y_2^2 & X_2 & 1 \\ X_3^2+Y_3^2 & X_3 & 1 \\ \end{vmatrix},\ d=-\begin{vmatrix} X_1^2+Y_1^2 & X_1 & Y_1 \\ X_2^2+Y_2^2 & X_2 & Y_2 \\ X_3^2+Y_3^2 & X_3 & Y_3 \\ \end{vmatrix} $$

Therefore, under the conditions $$a\not=0\quad\text{and}\quad \frac{b^2+c^2}{4a^2}-\frac da\gt 0$$ which are needed in order for such a circle to exist, we have

$$\begin{align}&\text{$A_4$ is inside the circle} \\\\&\iff\sqrt{\left(X_4-\left(-\frac{b}{2a}\right)\right)^2+\left(Y_4-\left(-\frac{c}{2a}\right)\right)^2}\le \sqrt{\frac{b^2+c^2}{4a^2}-\frac da} \\\\&\iff \left(X_4+\frac{b}{2a}\right)^2+\left(Y_4+\frac{c}{2a}\right)^2\le \frac{b^2+c^2}{4a^2}-\frac da \\\\&\iff X_4^2+\frac{bX_4}{a}+Y_4^2+\frac{cY_4}{a}+\frac da\le 0 \\\\&\iff a^2X_4^2+abX_4+a^2Y_4^2+acY_4+ad\le 0\end{align}$$ The LHS of the last inequality is a sixth degree polynomial on $t$.

Answer 2

First, find the center and radius of the circle. Ways to do this are shown here:

Get the equation of a circle when given 3 points

Then, find the times that the line passes through the circle.

Using vectors, if the center is at $C$, the radius is $r$, and the line is $A+Bt$, you want $|A+Bt-C| =r$ or $|A+Bt-C|^2 =r^2$ or

$\begin{array}\\ r^2 &=\sum (A_i-C_i+tB_i)^2\\ &=\sum (D_i+tB_i)^2 \qquad D_i = A_i-C_i\\ &=\sum (D_i^2+2tD_iB_i+t^2B_i^2)\\ &=\sum D_i^2+2t\sum D_iB_i+t^2\sum B_i^2 \end{array} $

This is a quadratic in $t$ and can be solved by the standard formula.

The values of $t$ tell when the line intersects the circle.

In terms of dot products, the equation for $t$ is $r^2 =D\cdot D+2tD\cdot B+t^2B\cdot B $.