The differential equation of the family of ci | vedclass

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(B) Let $(h, 0)$ be the center of the circle and $r$ be the radius. The equation of the circle is $(x-h)^2 + y^2 = r^2$. Since the radius $r$ is not f

Vedclass · Accepted Answer

$y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2+1=0$

Vedclass · Answer

(B) Let $(h, 0)$ be the center of the circle and $r$ be the radius. The equation of the circle is $(x-h)^2 + y^2 = r^2$. Since the radius $r$ is not fixed,this family of circles has two parameters $h$ and $r$. However,for a specific radius $r$,the family is $(x-h)^2 + y^2 = r^2$. Differentiating with respect to $x$: $2(x-h) + 2y \frac{dy}{dx} = 0$,which gives $x-h = -y \frac{dy}{dx}$. Substituting this into the circle equation: $(-y \frac{dy}{dx})^2 + y^2 = r^2$,so $y^2 (\frac{dy}{dx})^2 + y^2 = r^2$. Differentiating again with respect to $x$: $2y \frac{dy}{dx} (\frac{dy}{dx})^2 + y^2 (2 \frac{dy}{dx} \frac{d^2y}{dx^2}) + 2y \frac{dy}{dx} = 0$. Dividing by $2y \frac{dy}{dx}$ (assuming $y 
eq 0$ and $\frac{dy}{dx} 
eq 0$): $(\frac{dy}{dx})^2 + y \frac{d^2y}{dx^2} + 1 = 0$.

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

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