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 having the general solution $y=c(x-c)^2$ ($c$ is an arbitrary constant) is

Form the differential equation of the family of ellipses having foci on the $y$-axis and centre at the origin.

If $l$ and $m$ are the order and degree of the differential equation of all straight lines at a constant distance of $P$ units from the origin,then $l m^2+l^2 m=$

If $m$ and $n$ are the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis, then $m n-m+n=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module