I have two circles whose centre co-ordinates are known. I also have a line joining the centres of these two circles. How can I find the co-ordinates of two points, one each on the circumference of either circles, where the line joining their centres intersects them at their circumference.
Note: The radius of both circles are known. Also, the circles can be anywhere in the Cartesian plane.
On
Let $C_1(x_1,y_1)$ and $C_2(x_2,y_2)$ be the centers and $R_1$ adn $R_2$ be the radii of of the circles.
The unit vector $$u=\frac {<x_2-x_1,y_2-y_1>}{\sqrt {(x_2-x_1)^2+ (y_2-y_1)^2}}$$ is the direction vector of the line joining the centers.
Thus points of intersections are respectively $C_1\pm R_1 u $ and $C_2 \pm R_2 u$
On
COMMENT.- You know the center points $P_1(x_1,y_1)$, $P_2(x_2,y_2)$ and the radius $r_1$, $r_2$
►Line $P_1P_2$: $y=mx+c \text{ where } m=\dfrac{y_2-y_1}{x_2-x_1}\text{ and } c=\dfrac{x_2y_1-x_1y_2}{x_2-x_1}$.
►The coordinates you are looking for are given by the two following quadratic systems $$\begin{cases} (x-x_1)^2+(y-y_1)^2=r_1^2\\y=mx+c\end{cases}$$
$$\begin{cases}(x-x_2)^2+(y-y_2)^2=r_2^2\\y=mx+c\end{cases}$$ each of them giving a quadratic equation by evident substitution of $y$.
Here, I would assume the circles are not overlapping.
There are many ways to solve this problem, but I think using vectors is the fastest way.
Let’s arbitrarily define an origin. Then, let the position vectors of circle 1 and 2 be $\overrightarrow{p_1}$ and $\overrightarrow{p_2}$ respectively.
Now define a unit vector $\hat{n}$ that points from center of center of circle 1 to center of circle 2: $$\hat n=\frac{\overrightarrow{p_1}-\overrightarrow{p_2}}{| \overrightarrow{p_1}-\overrightarrow{p_2}|}$$
Then the position vector of the required point on circle 1 is $$\overrightarrow{c_1}=r_1\hat{n}+ \overrightarrow{p_1}$$ where $r_1$ is the radius of circle 1.
Similarly, $$\overrightarrow{c_2}=r_2(-\hat{n})+ \overrightarrow{p_2}$$