Find the orthogonal of the family of circles `x^(2) + y^(2) = 2ax` each of which touches the y-axis at origin.

Explanation

`x^(2) + y^(2) = 2ax`
Differentiating w.r.t. x, we have ... (i)
`2x + 2y(dy)/(dx) = 2a`
`rArr x + y(dy)/(dx) = a` ...(ii)
Eliminating a between (i) and (ii), we get
`x^(2) + y^(2) = 2(x+y(dy)/(dx))x`
`= 2x^(2) 2xy (dy)/(dx)`
`rArr y^(2) - 2xy(dy)/(dx) - x^(2) = 0` ....(iii)
Replacing `(dy)/(dx)` by `- (dx)/(dy)` to get the differential equation corresponding to the trajectory orthogonal to (i), we have
`y^(2) + 2xy(dx)/(dy) - x^(2) = 0`
`rArr x^(2) - 2xy(dx)/(dy) - y^(2) = 0` ...(iv)
Although (iv) can be solved the substitution y = vx, the differential equation being homogeneous, `(viz, (dx)/(dy) = (x^(2) - y^(2))/(2xy))`, we observe that (iv) is same as (iii) with x and y interchanged.
Hence the solution of (iv) is obtained by changing x to y in (i) i.e.,
`y^(2) + x^(2) = 2by`
which is another set of circles, each of which touches the x-axis at the origin.