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Question

(B) Let $(h, 8)$ be the centre of the circle.Since the circle touches the $X$-axis,its radius is $r = 8$.The equation of the circle is $(x-h)^{2} + (y

Vedclass · Accepted Answer

$(y-8)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64$

Vedclass · Answer

(B) Let $(h, 8)$ be the centre of the circle.Since the circle touches the $X$-axis,its radius is $r = 8$.The equation of the circle is $(x-h)^{2} + (y-8)^{2} = 8^{2} = 64$ ... $(1)$Differentiating equation $(1)$ with respect to $x$,we get:$2(x-h) + 2(y-8) \frac{d y}{d x} = 0$$(x-h) = -(y-8) \frac{d y}{d x}$Substituting the value of $(x-h)$ into equation $(1)$:$[-(y-8) \frac{d y}{d x}]^{2} + (y-8)^{2} = 64$$(y-8)^{2} (\frac{d y}{d x})^{2} + (y-8)^{2} = 64$$(y-8)^{2} [1 + (\frac{d y}{d x})^{2}] = 64$Thus,the correct option is $B$.

The differential equation of the circles having their centres on the line $y=8$ and touching the $X$-axis is

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