Question

Form the differential equation of all circles which pass through the origin and whose centers lie on X-axis.

Answer 1

Let C(h, 0) be the center of the circle which passes through the origin. Then the radius of the circle is h.

∴ equation of the circle is (x - h)² + (y - 0)² = h² 

∴ x² - 2hx + h + y² = h²

∴ x² + h² = 2hx        ....(1)

Differentiating both sides w.r.t. x, we get

`"2x" + "2y" "dy"/"dx" = "2h"`

Substituting the value of 2h in equation (1), we get

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

∴ `"x"^2 + "y"^2 = 2"x"^2 + 2"xy" "dy"/"dx"`

∴ `"2xy" "dy"/"dx" + "x"^2 - "y"^2 = 0`

This is the required D.E.