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I would like to determine the point $C$ in this image:

enter image description here

(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.

Zev Chonoles
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hotate17
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  • Welcome to math.SE. You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Feb 22 '13 at 08:21

2 Answers2

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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 $$x=x_0+\frac{r(x_1-x_0)}{\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}}.$$

Zev Chonoles
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Hint: Assuming the coordinates of unknown point are $(X,Y)$, solve these two equations simultaneously: $$(Y-y_0)^2+(X-x_0)^2=r^2,~~~Y=\frac{y_1-y_0}{x_1-x_0}(X-x_0)-y_0$$

Mikasa
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