The differential equation of all circles havi | vedclass

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

Question

(C) The equation of a circle with center $(h, 0)$ on the $x$-axis and radius $h$ (since it touches the $y$-axis at the origin) is $(x - h)^2 + y^2 = h

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equation of a circle with center $(h, 0)$ on the $x$-axis and radius $h$ (since it touches the $y$-axis at the origin) is $(x - h)^2 + y^2 = h^2$.Expanding this,we get $x^2 - 2xh + h^2 + y^2 = h^2$,which simplifies to $x^2 + y^2 = 2xh$.To eliminate the arbitrary constant $h$,we differentiate with respect to $x$:$2x + 2y \frac{dy}{dx} = 2h$.Substituting $h = x + y \frac{dy}{dx}$ into the original equation $x^2 + y^2 = 2x(h)$:$x^2 + y^2 = 2x(x + y \frac{dy}{dx})$.$x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$.$y^2 - x^2 = 2xy \frac{dy}{dx}$.Thus,$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

The differential equation of all circles having center on the $x$-axis and touching the $y$-axis at the origin is:

Similar Questions

The differential equation whose solution is $y = c_1 \cos ax + c_2 \sin ax$ is (where $c_1, c_2$ are arbitrary constants):

The differential equation for which $\sqrt{1+y^2}=C x e^{\tan ^{-1} x}$ is the general solution, is

The differential equation,having general solution as $A x^2+B y^2=1$,where $A$ and $B$ are arbitrary constants,is

The differential equation of the family of curves $v = \frac{A}{r} + B$,where $A$ and $B$ are arbitrary constants,is

Let $c_1, c_2, c_3, c_4$ be arbitrary constants. The order of the differential equation,corresponding to $y=c_1 e^x+c_2 e^{\log _{e} x}+c_3 \sin ^2 x-c_4\left(\cos ^2 x-1\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module