The differential equation of all circles pass | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $(x - h)^2 + y^2 = h^2$,where $h$ is the $

Vedclass · Accepted Answer

$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$

Vedclass · Answer

(D) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $(x - h)^2 + y^2 = h^2$,where $h$ is the $x$-coordinate of the center.Expanding this,we get $x^2 - 2hx + h^2 + y^2 = h^2$,which simplifies to $x^2 + y^2 - 2hx = 0$ ..... $(i)$.Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2h = 0$.Dividing by $2$,we get $x + y \frac{dy}{dx} - h = 0$,so $h = x + y \frac{dy}{dx}$.Substituting the value of $h$ into equation $(i)$,we get $x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0$.$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.$y^2 - x^2 - 2xy \frac{dy}{dx} = 0$.Rearranging the terms,we get $2xy \frac{dy}{dx} = y^2 - x^2$,which gives $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

The differential equation of all circles passing through the origin and having their centers on the $x$-axis is

Similar Questions

If $y = c{e^{{{\sin }^{ - 1}}x}}$,then the corresponding differential equation is:

The differential equation of the family of circles whose center lies on the $x$-axis is:

The differential equation of all parabolas each of which has a latus rectum $4a$ and whose axes are parallel to the $x$-axis is:

The family of curves $y = e^{a \sin x}$, where '$a$' is an arbitrary constant, is represented by the differential equation:

The differential equation of the family of curves represented by the equation ${x^2} + {y^2} = {a^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module