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

The Correct Option is A

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