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(D) The center of the circle is $(h, 5)$ and it touches the $X$-axis,so its radius $r$ is equal to the absolute value of the $y$-coordinate of the cen

Vedclass · Accepted Answer

$(5-y)^2 \left(\frac{dy}{dx}\right)^2 + y^2 - 10y = 0$

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(D) The center of the circle is $(h, 5)$ and it touches the $X$-axis,so its radius $r$ is equal to the absolute value of the $y$-coordinate of the center,which is $r = 5$.The equation of the circle is $(x-h)^2 + (y-5)^2 = 5^2$.$(x-h)^2 + (y-5)^2 = 25$.Differentiating with respect to $x$,we get $2(x-h) + 2(y-5) \frac{dy}{dx} = 0$.Thus,$(x-h) = -(y-5) \frac{dy}{dx}$.Substituting this into the circle equation: $[-(y-5) \frac{dy}{dx}]^2 + (y-5)^2 = 25$.$(y-5)^2 \left(\frac{dy}{dx}\right)^2 + (y-5)^2 = 25$.$(y-5)^2 \left(\frac{dy}{dx}\right)^2 + y^2 - 10y + 25 = 25$.$(y-5)^2 \left(\frac{dy}{dx}\right)^2 + y^2 - 10y = 0$.

The differential equation of all circles having their centres on the line $y=5$ and touching the $X$-axis is ......

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