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(C) The equation of a family of circles with constant radius $r$ and center $(a, b)$ is $(x-a)^2 + (y-b)^2 = r^2$. Differentiating with respect to $x$

Vedclass · Accepted Answer

$r^2\left(y^{\prime \prime}\right)^2=\left[1+\left(y^{\prime}\right)^2\right]^3$

Vedclass · Answer

(C) The equation of a family of circles with constant radius $r$ and center $(a, b)$ is $(x-a)^2 + (y-b)^2 = r^2$. Differentiating with respect to $x$: $2(x-a) + 2(y-b)y' = 0 \Rightarrow x-a = -(y-b)y'$. Substituting this into the circle equation: $(y-b)^2 (y')^2 + (y-b)^2 = r^2 \Rightarrow (y-b)^2 [1 + (y')^2] = r^2 \Rightarrow y-b = \frac{\pm r}{\sqrt{1+(y')^2}}$. Differentiating $x-a = -(y-b)y'$ again: $1 = -[y'' (y-b) + (y')^2]$. Substituting $y-b$: $1 + (y')^2 = -y'' \left( \frac{\pm r}{\sqrt{1+(y')^2}} \right)$. Rearranging: $1 + (y')^2 = \mp \frac{r y''}{\sqrt{1+(y')^2}} \Rightarrow (1+(y')^2)^{3/2} = \mp r y''$. Squaring both sides: $(1+(y')^2)^3 = r^2 (y'')^2$.

The differential equation representing the family of circles of constant radius $r$ is

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