Question - Coordinate Geometry (Circles)

I have been studying circles and came with this in my book :

The equation of the chord of the circle $x^2 + y^2 = a^2$ joining $(a\cos \alpha , a\sin \alpha) $ and $(a\cos \beta, a\sin \beta)$ is :

$x \cos (\dfrac{\alpha + \beta}{2}) + y\sin (\dfrac{\alpha + \beta}{2}) = a \cos (\dfrac{\alpha - \beta}{2}) $

Question

Question - Coordinate Geometry (Circles)

I have been studying circles and came with this in my book : 

The equation of the chord of the circle $x^2 + y^2 = a^2$ joining $(a\cos \alpha , a\sin \alpha) $ and $(a\cos \beta, a\sin \beta)$ is :

$x \cos (\dfrac{\alpha + \beta}{2}) + y\sin (\dfrac{\alpha + \beta}{2}) = a \cos (\dfrac{\alpha - \beta}{2}) $

mathslover · Answer

How can I prove the above equation?

accessdenied · Answer

We might start by finding the slope of the line joining those two points and setting up an equation using point-slope form:
   y - y0 = m (x - x0)
It probably won't look very pretty yet so we would need to do some hairy simplification work to get to that equation.

mathslover · Answer

$a\sin \beta - a\sin \alpha = m ( a\cos \beta - a \cos \alpha)  \ 
m = \dfrac{a\sin \beta - a\sin \alpha }{a\cos \beta - a \cos \alpha} $

accessdenied · Answer

Oh, we only use one of the points in point-slope.  We want to leave the plain x and y so that in the end we have an equation with x and y.
$ y - \color{blue}{y_0} = m \left( x - \color{blue}{x_0} \right) $

mathslover · Answer

Oh okay, fine! 
 So, shall I take ($x_0 , y_0$ ) $\equiv$ ($a \cos \alpha , a\sin \alpha$) or ($a \cos \beta , a \sin \beta$ )

accessdenied · Answer

I believe both will work out in the end, so either choice should be fine.  I'd go with the first one.

mathslover · Answer

Okay.. Great! I will work with your choice.

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