Find coordinates of unknown point given two other points and their distances There are two known points in the coordinate plane $A(x_1,y_1)$ and $B(x_2,y_2)$. The coordinates of point $C(x_0,y_0)$ are to be found, given that the distance between $A$ and $C$ is $a$ and distance between $B$ and $C$ is $b$. Also the angle between the lines $AC$ and $BC$ is $\theta$. How can $x_0$ and $y_0$ be expressed in terms of the known information? 
 A: HINT
The point can be found as the intersection of


*

*circle centered in $A$ with radius $a$: $(x-x_1)^2+(y-y_1)^2=a^2$

*circle centered in $B$ with radius $b$: $(x-x_2)^2+(y-y_2)^2=b^2$
that is


*

*$(x_0-x_1)^2+(y_0-y_1)^2-a^2=(x_0-x_2)^2+(y_0-y_2)^2-b^2$


from here we can find a linear equation in $x_0$ and $y_0$. Then by the intersection with one of the circle we can find out the point C (we may have $0,1,2$ or infinitely many solutions).
A: You need the formulas $$a=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$$
$$b=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2}$$ and
$$\tan(\theta)=\vert{\frac{m_1-m_2}{1+m_1m_2}\vert}$$
A: 
Consider points $A,B,C$
as vertices of $\triangle ABC$
with $|AC|=a$, $|BC|=b$ and $\angle ACB=\theta$. 
Let $c=|AB|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
Note also that $\theta$ 
can be found from $a,b,x_1,y_1,x_2,y_2$.
The sought point $C$ can be found by rotating
the point 
\begin{align}
C_0&=A+(B-A)\cdot\frac{a}c
\end{align}
around the point $A$
by the angle $\pm\alpha$.
There is no solution, if the numbers
$a,b,c$ does not form a valid triangle,
otherwise
there are two solutions, (unless $\sin\alpha=0$):
\begin{align}
x_0&=
x_1+\frac{a}c\,(x_2-x_1)\cos\alpha-\frac{a}c\,(y_2-y_1)\sin\alpha
,\\
y_0&= y_1+\frac{a}c\,(y_2-y_1)\cos\alpha+\frac{a}c\,(x_2-x_1)\sin\alpha
\end{align}
and
\begin{align}
x_0&=
x_1+\frac{a}c\,(x_2-x_1)\cos\alpha+\frac{a}c\,(y_2-y_1)\sin\alpha
,\\
y_0&= y_1+\frac{a}c\,(y_2-y_1)\cos\alpha-\frac{a}c\,(x_2-x_1)\sin\alpha
,
\end{align}
where
\begin{align}
\cos\alpha&=\frac{a^2+c^2-b^2}{2ac}
,\\
\sin\alpha&=\sqrt{1-\cos^2\alpha}
.
\end{align}
