Question

The differential equation of all circles having their centres on the line \(y=5\) and touching ( X -axis) is \(\qquad\)

• A. \((5-y)\frac{\mathrm{d}y}{\text{ }\mathrm{d}x}+y^{2}-10y=0\)
• B. \((5-y)^{2}\frac{\text{ }\mathrm{d}^{2}y}{\text{ }\mathrm{d}x^{2}}+y^{2}-10y=0\)
• C. \((5-y)\frac{\mathrm{d}y}{\text{ }\mathrm{d}x}+y-10=0\)
• D. \((5-y)^{2}(\frac{\text{ }\mathrm{d}y}{\text{ }\mathrm{d}x})^{2}+y^{2}-10y=0\)

Hint:

The radius is fixed at 5, so there is one arbitrary constant, the x coordinate of the centre. Eliminate it by differentiating once.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
A circle with centre on \(y=5\) that touches the X-axis has centre \((h,5)\) and radius 5.
Its equation is \((x-h)^2+(y-5)^2=25\).

Step 2: Differentiate
\(2(x-h)+2(y-5)\dfrac{dy}{dx}=0\Rightarrow x-h=-(y-5)\dfrac{dy}{dx}\)

Step 3: Eliminate h
Substitute into the circle: \((y-5)^2\left(\dfrac{dy}{dx}\right)^2+(y-5)^2=25\).

Step 4: Simplify
\((5-y)^2\left(\dfrac{dy}{dx}\right)^2+y^2-10y+25-25=0\)
\((5-y)^2\left(\dfrac{dy}{dx}\right)^2+y^2-10y=0\)

Final Answer:
The differential equation is the first-order second-degree equation in option D. \[ \boxed{\text{(D) }(5-y)^{2}\left(\dfrac{dy}{dx}\right)^{2}+y^{2}-10y=0} \]