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(B) Let the centre of the circle on the $y$-axis be $(0, b)$.The equation of the family of circles with centre $(0, b)$ and radius $3$ is:$x^2 + (y -

Vedclass · Accepted Answer

$(x^2 - 9)(y')^2 + x^2 = 0$

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(B) Let the centre of the circle on the $y$-axis be $(0, b)$.The equation of the family of circles with centre $(0, b)$ and radius $3$ is:$x^2 + (y - b)^2 = 3^2$$x^2 + (y - b)^2 = 9$  --- $(1)$Differentiating equation $(1)$ with respect to $x$,we get:$2x + 2(y - b) \cdot y' = 0$$(y - b) \cdot y' = -x$$(y - b) = -\frac{x}{y'}$Substituting the value of $(y - b)$ in equation $(1)$:$x^2 + \left(-\frac{x}{y'}\right)^2 = 9$$x^2 + \frac{x^2}{(y')^2} = 9$Multiplying by $(y')^2$:$x^2(y')^2 + x^2 = 9(y')^2$$x^2(y')^2 - 9(y')^2 + x^2 = 0$$(x^2 - 9)(y')^2 + x^2 = 0$

Form the differential equation of the family of circles having centre on the $y$-axis and radius $3$ units.

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