The differential equation of the family of ci | vedclass

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Question

(A) The equation of the family of circles with fixed radius $r$ and center $(h, 3)$ is given by:$(x-h)^2 + (y-3)^2 = r^2$  --- $(1)$where $h$ is an ar

Vedclass · Accepted Answer

$1+\left(\frac{dy}{dx}\right)^2=\frac{r^2}{(y-3)^2}$

Vedclass · Answer

(A) The equation of the family of circles with fixed radius $r$ and center $(h, 3)$ is given by:$(x-h)^2 + (y-3)^2 = r^2$  --- $(1)$where $h$ is an arbitrary parameter.Differentiating both sides with respect to $x$:$2(x-h) + 2(y-3)\frac{dy}{dx} = 0$$x-h = -(y-3)\frac{dy}{dx}$Substituting $(x-h)$ into equation $(1)$:$[-(y-3)\frac{dy}{dx}]^2 + (y-3)^2 = r^2$$(y-3)^2 \left(\frac{dy}{dx}\right)^2 + (y-3)^2 = r^2$Dividing both sides by $(y-3)^2$:$\left(\frac{dy}{dx}\right)^2 + 1 = \frac{r^2}{(y-3)^2}$

The differential equation of the family of circles with fixed radius $r$ units and centre on the line $y=3$ is

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