Question

The differential equation of the family of circles passing through the points (0, 2) and (0, –2) is

• A. $\begin{array}{l}\ 2xy\frac{dy}{dx}+\left(x^2-y^2+4\right)=0 \end{array}$
• B. $\begin{array}{l} \ 2xy\frac{dy}{dx}+\left(x^2+y^2-4\right)=0\end{array}$
• C. $\begin{array}{l} \ 2xy\frac{dy}{dx}+\left(y^2-x^2+4\right)=0\end{array}$
• D. $\begin{array}{l} \ 2xy\frac{dy}{dx}-\left(x^2-y^2+4\right)=0\end{array}$

Answer 1

Answer: (A)

$\begin{array}{l}\ 2xy\frac{dy}{dx}+\left(x^2-y^2+4\right)=0 \end{array}$

Answer 2

Let the equation of the family of circles be

$x^2+y^2+2gx+2fy+c=0$

It passes through (a, 0) and (− a, 0). Therefore,

$a^2+2ag+c=0$ and $a^2−2ag+c=0$

Solving these two equations, we get $c=−a^2$ and $g=0$

Substituting the values of c and gin (i), we get

$x^2+y^2+2fy−a^2=0$

It is a one parameter family of circles.

Differentiating with respect to x, we get

$2x+2yy_1+2fy_1=0⇒f=−(\frac{x+yy_1}{y_1})$

Substituting the value off in (ii), we get

$y_1(y^2−x^2+a^2)+2xy=0$

This is the required differential equation