Question

Find orthogonal trajectory of x2+y2+2gx+c=0

Answer 1

x² + y² + 2gx + c = 0 ----------(1)

Differentiating w.r.t. x, we get:

2x + 2y(dy/dx) + 2g = 0

==> 2g = -( 2x + 2y(dy/dx) )

Substituting this value in eq(1), we get:

x² + y² - x[ 2x + 2y(dy/dx) ] + c = 0 ----------(2)

x² + y² - 2x² - 2xy(dy/dx) + c = 0

Replacing dy/dx by -dx/dy -->

x² + y² - 2x² + 2xy(dx/dy) + c = 0

y² - x² + 2xy(dx/dy) + c = 0

y²dy- x²dy+ 2xydx + cdy = 0

2xydx + ( y²- x² + c )dy = 0 ----------(3)

Eq(3) is of type 3 equation:

Thus,

Here M = 2xy and N =  y²- x² + c

dN/dx = -2x

dM/dy = 2x

₍N_x - M_y ) / M = (-2x - 2x) / 2xy

= -4x / 2xy

= -2 / y

I.F. = e^{∫-2/y dy} = e^-2logy = e^{log y^(-2)} = y^-2 = 1/y²

Multiplying I.F. with eq(3)

(1/y2) [2xydx + ( y²- x² + c )dy  = 0 ]

2x/y dx + (1 - (x/y)² +cy^-2) dy = 0 ---------(4)

This is now an exact first order differential equation, so the solution of this equation is 

∫_{y constant} Mdx + ∫(terms not containing x) = c

In eq(4), M = 2x/y

∫2x/y dx + ∫(1+cy) dy = c

x²/y + y -c/y = c

x² + y² - c = cy

x² + y² -cy - c = 0

The general equation of circle is x² + y² +2gx - 2fy + c = 0

Thus, x² + y² - (-2f)y - c = 0

x² + y² + 2fy - c = 0

This is the required orthognal trajectory of the eq(1).