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

The differential equation for the family of curves ${x^2} + {y^2} - 2ay = 0,$ where $a$ is an arbitrary constant,is

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

The differential equation of the circles having their centres on the line $y=8$ and touching the $X$-axis is

Verify that the given function $x^{2}=2 y^{2} \log y$ is a solution of the corresponding differential equation $(x^{2}+y^{2}) \frac{dy}{dx}-xy=0$.

Verify that the function $y=e^{-3x}$ is a solution of the differential equation $\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-6y=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module