circlesgeometry
Probably simple to solve but I'm a bit stuck.
I am given two lines that are tangent to a circle and the circle must go through $P_1$ (which is the end of Line 1) and $P_2$ (which is the end of Line 2).
How do I calculate the Center Point of that circle? With given lines and points it should be only one solution.
An identity that might prove useful in this problem is $$ \cot\left(\frac{\theta}{2}\right)=\frac{\sin(\theta)}{1-\cos(\theta)}=\frac{1+\cos(\theta)}{\sin(\theta)}\tag{1} $$ In $\mathbb{R}^3$, one usually uses the cross product to compute the sine of the angle between two vectors. However, one can use a two-dimensional analog of the cross product to do the same thing in $\mathbb{R}^2$.
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In the diagram above, $(x,y)\perp(y,-x)$ and so the $\color{#FF0000}{\text{red angle}}$ is complementary to the $\color{#00A000}{\text{green angle}}$. Thus, $$ \begin{align} \sin(\color{#FF0000}{\text{red angle}}) &=\cos(\color{#00A000}{\text{green angle}})\\[6pt] &=\frac{(u,v)\cdot(y,-x)}{|(u,v)||(y,-x)|}\\[6pt] &=\frac{uy-vx}{|(u,v)||(x,y)|}\tag{2} \end{align} $$ $uy-vx$ is the normal component of $(u,v,0)\times(x,y,0)$; thus, it is a two dimensional analog of the cross product, and we will denote it as $(u,v)\times(x,y)=uy-vx$.
Let $u_a=\frac{Q-P}{|Q-P|}$ and $u_b=\frac{R-P}{|R-P|}$, then since $$ |A-P|=|B-P|=r\cot\left(\frac{\theta}{2}\right) $$ we get $$ \begin{align} A &=P+ru_a\cot\left(\frac{\theta}{2}\right)\\ &=P+ru_a\frac{1+u_a\cdot u_b}{|u_a\times u_b|} \end{align} $$ and $$ \begin{align} B &=P+ru_b\cot\left(\frac{\theta}{2}\right)\\ &=P+ru_b\frac{1+u_a\cdot u_b}{|u_a\times u_b|} \end{align} $$ where we take the absolute value of $u_a\times u_b$ so that the circle is in the direction of $u_a$ and $u_b$.
You haven't made a mistake. Here's an image of your arrangement. There are two solutions. The red line is the line from P1 to P2. The two tangent points are indeed measured as your solved x values.
Best Answer
Following the comment by martini: since every radius of circle is perpendicular to the corresponding tangent line, the center $O$ must be such that $OP_1\perp \ell_1$ and $OP_2\perp \ell_2$. This already determines $O$ as the point of intersection of the perpendiculars to $OP_j$ passing through $P_j$, $j=1,2$.
The solution is unique, if it exists; but it does not exist when $|OP_1|\ne |OP_2|$. (The problem is overdetermined, as Hagen von Eitzen said.)