Question

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

Answer 1

The equation of the family of circles in the second quadrant and touching the co-ordinate axes is
\[\left( x + a \right)^2 + \left( y - a \right)^2 = a^2 \]
\[ \Rightarrow x^2 + 2ax + a^2 + y^2 - 2ay + a^2 = a^2 \]
\[ \Rightarrow x^2 + 2ax + y^2 - 2ay + a^2 = 0 ..........(1)\]
where `a` is a parameter.
As this equation contains one parameter, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
\[2x + 2a + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0\]
\[ \Rightarrow x + y\frac{dy}{dx} + a - a\frac{dy}{dx} = 0\]
\[ \Rightarrow \left( x + y\frac{dy}{dx} \right) + a\left( 1 - \frac{dy}{dx} \right) = 0\]
\[ \Rightarrow a = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx} - 1} ..........(2)\]
From (1) and (2), we get 
\[x^2 + 2x\left( \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx} - 1} \right) + y^2 - 2y\left( \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx} - 1} \right) + \left( \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx} - 1} \right)^2 = 0\]
\[ \Rightarrow x^2 \left( \frac{dy}{dx} - 1 \right)^2 + 2x\left( x + y\frac{dy}{dx} \right)\left( \frac{dy}{dx} - 1 \right) + y^2 \left( \frac{dy}{dx} - 1 \right)^2 - 2y\left( x + y\frac{dy}{dx} \right)\left( \frac{dy}{dx} - 1 \right) + \left( x + y\frac{dy}{dx} \right)^2 = 0\]
\[ \Rightarrow x^2 \left( \frac{dy}{dx} \right)^2 - 2 x^2 \left( \frac{dy}{dx} \right) + x^2 + 2x\left[ x\frac{dy}{dx} - x + y \left( \frac{dy}{dx} \right)^2 - y\frac{dy}{dx} \right] + y^2 \left[ \left( \frac{dy}{dx} \right)^2 - 2\frac{dy}{dx} + 1 \right] - 2y\left[ x\frac{dy}{dx} - x + y \left( \frac{dy}{dx} \right)^2 - y\frac{dy}{dx} \right] + x^2 + 2xy\frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2 = 0\]
\[ \Rightarrow x^2 + 2xy\frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2 = x^2 + 2xy + y^2 + \left( x^2 + 2xy + y^2 \right) \left( \frac{dy}{dx} \right)^2 \]
\[ \Rightarrow \left( x + y\frac{dy}{dx} \right)^2 = \left( x + y \right)^2 \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] \]
It is the required differential equation.