How to find $P_1$ in $(x,y)$ form

Question

From following diagram, $A_1$ is center of circle of radius $r$. All distances are in coordinate system $(x,y)$. Distance from $A_1P_2$ is known. Distance $A_1,A_2,A_3$ is also known from origin. I want to find point $P_1$.

I appreciate any help in this regard.

Thanks.

Answer 1

The following are equal. They have the same direction and unit length: $$ \frac{P_1-A_1}{r}=\frac{P_2-A_1}{|P_2-A_1|} $$ Therefore, assuming we know the coordinates of $P_2$ and $A_1$, $$ P_1=A_1+r\frac{P_2-A_1}{|P_2-A_1|} $$

Answer 2

The OP clarified in chat that he knows coordinates of $P_2$ and simply wants to compute the coordinates of $P_1$, i.e., the intersection of the circle around $A_1$ and the line going through $A_1$ and $P_2$. I hope they will also clarify this in the question.

Suppose that $A_1$ has coordinates $(x_A,y_A)$ and $P_2$ has coordinates $(x_P,y_P)$. Then the line going through $A_1$ and $P_2$ is $$\frac{y-y_A}{x-x_A}=\frac{y_P-y_A}{x_P-x_A}$$ or $$(x_P-x_A)(y-y_A)=(x-x_A)(y_P-y_A).$$

You also want $$(x-x_A)^2+(y-y_A)^2=r^2.$$

Let us denote $a=x-x_A$, $b=y-y_A$. (Which is the same as translating the origin of the coordinate system to the point $A$.) Then you have two equations $$(x_P-x_A)a=(y_P-y_A)b\\ a^2+b^2=r^2.$$ From the first equation you get $b=\frac{x_P-x_A}{y_P-y_A}a$. (You should think about the special case $y_P=y_A$ separately; you cannot divide by zero.)

If you plug this into the second equation you get $$b^2\left(1+\frac{(x_P-x_A)^2}{(y_P-y_A)^2}\right)=r^2.$$ You can solve for $b$. If you know $b$, you can compute $a$. Once you have $a$ and $b$ you can compute $x$ and $y$ easily.

You will get two solutions, since the line and the circle intersect in two points.

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