solve `(xdx+ydy)/(xdy - ydx) =sqrt((a^2 - x^2 - y^2)/ (x^2 + y^2))`

Explanation

Since `x^(2) + y^(2)` appears as a combination, we make the substitution
x = cos that `rArr dx = -r sin theta d theta + cos theta dr`
`y = r sin theta rArr dy = r cos theta d theta + sin theta dr` (Note that r and `theta` are both variables)
`xdx + ydy = r cos` that `(-r sin theta d theta + cos theta dr) + r sin theta( r cos theta d theta + sin theta d r)`
`= -r^(2) sin theta cos theta d theta + r cos^(2) theta dr + r^(2) sin theta cos theta d theta + r sin^(2) theta dr`
= rdr ...(A)
Again,
`xdy - ydx = r cos theta(r cos theta d theta + sin theta dr) - r sin theta(-r sin theta d theta + cos theta dr)`
`= r^(2) cos^(2) theta d theta + r cos theta sin theta dr + r^(2) sin^(2) theta d theta - r sin theta cos theta dr`
`= r^(2) d theta` ....(B)
(A) and (B) could also be derived in another elegant way
`x = r cos theta, y = r sin theta` given `r^(2) = x^(2) + y^(2)`
`rArr 2r dr = 2xdx + 2ydy :. rdr = xdx + ydx`
Again,
`tan theta = (y)/(x) rArr sec^(2).theta d theta = (xdy - y dx)/(x^(2))`
`rArr xdy - ydx = x^(2) sec^(2) theta d theta`
`= (r^(2) cos^(2) theta) sec^(2) theta d theta = r^(2) d theta`
`:. xdy - ydx = r^(2) dtheta`
The given equation now transform to
`(xdx + ydy)/(xdx - ydy) = sqrt((a^(2).(x^(2) + y^(2)))/(x^(2) + y^(2)))`
`rArr (rdr)/(r^(2) d theta) = sqrt((a^(2)-r^(2))/(r^(2))) rArr d theta = (dr)/(sqrt(a^(2) - r^(2)))`
Integrating yields `k + theta = sin^(-1).(r)/(a)` (k is an arbitrary constant.)
`rArr r = a sin (k + theta) rArr x^(2) + y^(2) = a sin (k + tan^(-1).(y)/(x))`