circles
I would like to determine the point $C$ in this image:
(assume I have radius value).
After the hours of research and refreshing some memories from school days, I've got:
Please assume…
Point: (xCoordinate, yCoordinate):
A (center): $(x_0,y_0)$
B: $(x_1,y_1) $
Plugging them into equation of circle
$$x^2 + \left(y_0 + \left( \frac{y_1-y_0}{x_1-x_0}\right) (x-x_0)\right)^2 = r^2$$
Now how can I solve for $x$? I would like to have the equation that starts with "$x=$".
Thanks a bunch in advance.
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 $u_b$.
Let's call the center of the circle: $P_1 = (x_1, y_1).\;$
Let $ P_2 = (x_2, y_2)$ be a point on circle. Then:
$r$: radius of the circle = distance between points $P_1$ and $P_2$, where
$$r = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}$$
Any point $(x_i, y_i)$ satisfying the equation $(x_i - x_1)^2 + (y_i - y_1)^2 = r^2$ also lies on this circle.
Best Answer
Because the point $C=(x,y)$ is lies on a line through $(x_0,y_0)$ and $(x_1,y_1)$, it is of the form $$(x,y)=(x_0+t(x_1-x_0),y_0+t(y_1-y_0))$$ for some real number $t$. In order for $C$ to lie on the circle, we must have that $$(t(x_1-x_0))^2+(t(y_1-y_0))^2=r^2,$$ and hence $$t=\frac{r}{\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}}.$$ Thus, the $x$-coordinate of $C$ is $$x=x_0+\frac{r(x_1-x_0)}{\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}}.$$