# By substitution or otherwise verify that the circular path satisfies the equation

A beam of light travelling between the points (0, $$y_0$$) and $$(x_1, y_1)$$ in the (x, y) plane will travel along the curve y(x) where y' = dy/dx.

I have worked out that the path y(x) must satisfy the equation

$$y\sqrt{1+y'^2}=\frac{1}{c}$$

the beam of light is travelling within the infinitely wide strip between y=$$y_0$$ and y=$$2y_0$$ where $$y_0$$>0

By substitution or otherwise, I must verify that the circular path satisfies my previous equation.

$$y(x)^2+ (x-x_0)^2=\frac{1}{c^2}$$

I have tried making the equations equal each other but that has come up blank I think I have to find y'(x) when I try to differentiate y(x) all I get is y'(x)=-1 and that doesn't help very much either. all help would be greatly appreciated as this question has me really stumped. feel free to adapt tags as you see fit I wasn't sure what to put for this question

Square the first equation and set it equal to the second $$y^2+ (x-x_0)^2=y^2(1+(y')^2 ) \\[9pt] \iff(x-x_0)^2 = (yy')^2$$ Now implicitly differentiate the second equation
$$2yy' +2(x-x_0) = 0 \iff yy'=-(x-x_0)$$
• $y$ is the name of a function of $x$, $y(x)$ and $y$ mean the same thing.