Question

Find the differential equation of all circles passing through the origin and having their centres on the x-axis.

Answer 1

The general equation of a circle is given by x^2 + y^2 + 2gx + 2fy + c = 0. Since the circle passes through the origin and its center lies on the x-axis, the center of the circle is (-g, 0).

Differentiating the equation with respect to x, we have:

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

Simplifying this equation, we get:

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

Substituting g = -2x/(x^2 + y^2), we have:

2x + 2y(dy/dx) = 2x/(x^2 + y^2)

Cancelling the common factors of 2x, we obtain:

1 + y(dy/dx) = 1/(x^2 + y^2)

Rearranging the equation, we get:

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

Multiplying both sides by dx, we have:

ydy = dx/(x^2 + y^2) - dx

Hence, the differential equation for a circle passing through the origin and having its center on the x-axis is given by:

ydy = dx/(x^2 + y^2) - dx