Question

The differential equation of family of circles having centre on line $y=10$ and touching x-axis is

• A. $\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}+6y=0$
• B. $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$
• C. ${8\left(\frac{dy}{dx}\right)}^3-27y=0$
• D. ${(y-10)}^2{\left(\frac{dy}{dx}\right)}^2+y^2-20y=0$

Answer 1

The correct option is D ${(y-10)}^2{\left(\frac{dy}{dx}\right)}^2+y^2-20y=0$

The general equation for a circle can be written as $x^2+y^2+2gx+2fy+c=0$

The center lies on $y=10$, and so $f=-10.$

Also, distance of the center from x axis has to be equal to the radius.

$\Rightarrow f^2=g^2+f^2-c$

$\Rightarrow g^2=c$

The equation thus becomes $x^2+y^2+2gx-20y+g^2=0$

Differentiating w.r.t. x, we get $2x+2y\frac{dy}{dx}+2g-20\frac{dy}{dx}=0$

$\Rightarrow g=(10-y)\frac{dy}{dx}-x$

Resubstituting this value in the above equation, we get $x^2+y^2-20y+2x\left[\right(10-y)\frac{dy}{dx}-x]+{\left[\right(10-y)\frac{dy}{dx}-x]}^2$

$=x^2+y^2-20y+2x(10-y)\frac{dy}{dx}-2x^2+(10-y{)}^2{\left(\frac{dy}{dx}\right)}^2+x^2-2x(10-y)\frac{dy}{dx}$

$=y^2-20y+(10-y{)}^2{\left(\frac{dy}{dx}\right)}^2$