Finding the location of point P

Question

Is it possible to find the location of point $P$ such that the angles $\theta_1=\theta_2$ or $\alpha_1=\alpha_2$? I know only the locations of $O$, $C_1$, and $C_2$.

Answer 1

If you are on a coordinate system, consider $P=(a,b)$ and find the equation of the line $\overline{OP}$. The point $S$ of intersection of lines $\overline{OP}$ and $\overline{C_1C_2}$ is computible. Finally note that $PS$ is the bisector of angle $C_1PC_2$, thus we have $$\frac{PC_1}{PC_2}=\frac{C_1S}{SC_2}.$$

Answer 2

Here's a geometric construction for $P$. On $OC_1$ as base construct an isosceles triangle $OBC_1$ such that $\angle BC_1O=\angle BOC_1=\alpha-90°$. All points of the circle arc of center $B$ and passing through $O$ make with $OC_1$ an angle $\alpha$. Construct in the same way another isosceles triangle $OAC_2$ and an arc $OC_2$: point $P$ lies then at the intersection of those two arcs.

EDIT.

If the triangle is right-angled at $O$ it is quite easy to find the equation of the locus of point $P$. In that case we can choose the coordinates of the vertices as $$ O=(0,0),\quad C_1=(a,0),\quad C_2=(0,b) $$ and the coordinates of the circle centers as $$ A=(a/2,-ta/2),\quad B=(-tb/2,b/2), $$ where $t$ is a real parameter. The coordinates of $P$ turn out to be: $$ P=\left( {ab(1-t^2)(at+b)\over(at+b)^2+(bt+a)^2}, {ab(1-t^2)(bt+a)\over(at+b)^2+(bt+a)^2} \right) $$ and they thus define a parametric curve (shown in diagram below). One can also eliminate $t$ to find the implicit equation of the curve: $$ (x^2+y^2)(ay-bx)=ab(y^2-x^2). $$

Notice that only the points on the blue and red arcs of the curve satisfy the request $\angle C_1PO=\angle C_2PO$, while the points on the dashed arcs don't. The blue and red arcs are then the required locus.