Question

Form the differential equation of the family of circles touching the 𝑥-axis at origin.

Answer 1

According to question the given curve is
Since the circle touches the \(x\)-axis at origin.
So, the centre will lie on the \(y\)-axis. So, \(x\)-coordinate of center is \(0\) i.e., \(a=0\)
Centre =\((0,b)\) and radius \(=b\)

Let

\(\dfrac{dy}{dx}=y'\)

So, Equation of Circle

\((x-0)^2+(y-b)^2=b^2\)

\(x^2+(y-b)^2=b^2\)

\(x^2+y^2-2yb+b^2=b^2\)

\(x^2=Y^2-2yb=0\)

\(x^2+y^2=2yb\)

Differentiating both sides w.r.t. x
we get,

\(2x+2y\dfrac{dy}{dx}=2b.\dfrac{dy}{dx}\)

\(x+yy'=b.y'\)

\(\left[\dfrac{x+yy'}{y'}\right]=b\)

\(x^2+y^2=2yb\)

putting

\(b=\left[\dfrac{x+yy'}{y'}\right]\)

\(x^2+y^2=2y\left[\dfrac{x+yy'}{y'}\right],\)

\((x^2+y^2)y'=2y(x+yy')\)

\((x^2+y^2)y'=2yx+2y^2y'\)

\(x^2y'+y^2y'=2yx+2y^2y'\)

\(x^2y'+y^2y'-2y^2y'=2yx\)

\(x^2y'-y^2y'=2yx\)

\(y'(x^2-y^2)=2xy\)

\(y'=\dfrac{2xy}{x^2-y^2}\)

Final Answer:
Hence, the required differential equation is

\(y'=\dfrac{2xy}{x^2-y^2}\)