Login/Sign Up
Home » The differential equation of the family of circles touching -axis at the origin is A) B) C) D)
The differential equation of the family of circles touching y-axis at the origin is
A) \left(x^{2}+y^{2}\right) \frac{d y}{d x}-2 x y=0
B) x^{2}-y^{2}+2 x y \frac{d y}{d x}=0
C) \left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y=0
D) \left(x^{2}+y^{2}\right) \frac{d y}{d x}+2 x y=0
Maths
The differential equation of the family of circles touching y-axis at the origin is
A) \left(x^{2}+y^{2}\right) \frac{d y}{d x}-2 x y=0
B) x^{2}-y^{2}+2 x y \frac{d y}{d x}=0
C) \left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y=0
D) \left(x^{2}+y^{2}\right) \frac{d y}{d x}+2 x y=0

Solutions:

3: (B)
As it is required circle touches y-axis at the origin.
\therefore Let us suppose Centre of the circle is d(a, 0) and radius is a
\therefore \quad Eq. of circle will be,
\begin{array}{l} (x-a)^{2}+(y-0)^{2}=a^{2} \\ \Rightarrow \quad x^{2}-2 a x+a^{2}+y^{2}=a^{2} \\ \Rightarrow \quad x^{2}+y^{2}-2 a x=0\dots \dots(i) \end{array}
By differentiating above eq. with respect to \mathrm{x}, we obtain
\begin{array}{l} 2 x+2 y \frac{d y}{d x}-2 a=0 \\ \therefore \quad 2 a=2 x+2 y \frac{d y}{d x}\dots \dots(ii) \end{array}
From equation(i) and equation(ii),
\begin{array}{l} x^{2}+y^{2}-\left(2 x+2 y \frac{d y}{d x}\right) x=0 \\ \Rightarrow \quad x^{2}+y^{2}-2 x^{2}-2 x y \frac{d y}{d x}=0 \\ \Rightarrow \quad x^{2}-y^{2}+2 x y \frac{d y}{d x}=0 \end{array}

i

More Material