The differential equation of all circles touc | vedclass

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Question

(B) The general equation of a circle touching the $Y$-axis at the origin $(0,0)$ and having its center on the $X$-axis is $(x-a)^2 + (y-0)^2 = a^2$,wh

Vedclass · Accepted Answer

$x^2-y^2+2xy \frac{dy}{dx}=0$

Vedclass · Answer

(B) The general equation of a circle touching the $Y$-axis at the origin $(0,0)$ and having its center on the $X$-axis is $(x-a)^2 + (y-0)^2 = a^2$,where $a$ is the radius of the circle.Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$.To eliminate the arbitrary constant $a$,we differentiate both sides with respect to $x$:$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(2ax)$$2x + 2y \frac{dy}{dx} = 2a$.Substituting the value of $2a = \frac{x^2+y^2}{x}$ into the differentiated equation:$2x + 2y \frac{dy}{dx} = \frac{x^2+y^2}{x}$.Multiplying both sides by $x$:$2x^2 + 2xy \frac{dy}{dx} = x^2 + y^2$.Rearranging the terms,we get $x^2 - y^2 + 2xy \frac{dy}{dx} = 0$.

The differential equation of all circles touching the $Y$-axis at the origin and having their center on the $X$-axis is:

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