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(B) The equation of a family of circles with radius $a$ and center $(h, k)$ is given by $(x - h)^2 + (y - k)^2 = a^2$. Since there are two arbitrary c

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$(1 + y_1^2)^3 = a^2 y_2^2$

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(B) The equation of a family of circles with radius $a$ and center $(h, k)$ is given by $(x - h)^2 + (y - k)^2 = a^2$. Since there are two arbitrary constants $h$ and $k$, we differentiate twice with respect to $x$. Differentiating with respect to $x$: $2(x - h) + 2(y - k)y_1 = 0$, which implies $(x - h) = -(y - k)y_1$. Differentiating again: $1 + y_1^2 + (y - k)y_2 = 0$, so $(y - k) = -\frac{1 + y_1^2}{y_2}$. Substituting $(y - k)$ back into the first derivative equation: $(x - h) = -(-\frac{1 + y_1^2}{y_2})y_1 = \frac{y_1(1 + y_1^2)}{y_2}$. Now substitute $(x - h)$ and $(y - k)$ into the original circle equation: $(\frac{y_1(1 + y_1^2)}{y_2})^2 + (-\frac{1 + y_1^2}{y_2})^2 = a^2$. This simplifies to $\frac{y_1^2(1 + y_1^2)^2}{y_2^2} + \frac{(1 + y_1^2)^2}{y_2^2} = a^2$. Factoring out $(1 + y_1^2)^2$: $\frac{(1 + y_1^2)^2 (y_1^2 + 1)}{y_2^2} = a^2$. Thus, $(1 + y_1^2)^3 = a^2 y_2^2$.

The differential equation of the family of all circles of radius $a$ is

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