Question

The differential equation of the family of circles passing through the fixed points $(a,0)$ and $(-a,0)$ is:

• A. $y_1(y^2-x^2)+2xy+a^2=0$
• B. $y_1{y}^2+xy+a^2{x}^2=0$
• C. $y_1(y^2-x^2+a^2)+2xy=0$
• D. none of these

Answer 1

Answer: (C)

$y_1(y^2-x^2+a^2)+2xy=0$

Let the equation of the circle be

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

As it passes through $(a,0)$ and $(-a,0)$

We have $a^2+2ga+c=0$ and $a^2-2ga+c=0$

Subtracting we obtain $g=0\Rightarrow c=-a^2$

Hence the equation of circle reduce to

$x^2+y^2+2fy-a^2=0$

Differentiating, we have

$x+y{y}_1+f{y}_1=0\Rightarrow f=\frac{-(x+y{y}_1)}{y_1}$

Thus we have $x^2+y^2-a^2-\frac{2x(x+y{y}_1)}{y_1}=0$

$\Rightarrow {-x}^2+y^2{-a}^2-\frac{2xy}{y_1}=0\Rightarrow y_1(y^2-x^2-a^2)+2xy=0$