Question

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

• A. $\frac{dy}{dx}(y^2-x^2)+2xy+a^2=0$
• B. $\frac{dy}{dx}{y}^2+xy+a^2{x}^2=0$
• C. $\frac{dy}{dx}(y^2-x^2+a^2)+2xy=0$
• D. $\frac{dy}{dx}(y^2+x^2-a^2)+2xy=0$

Answer 1

Answer: (C)

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

Equation of circle with center $(h,k)$ is :

$(x-h{)}^2+(y-k{)}^2=r^2$

It passes through $(a,0)$ and $(-a,0)$

$\Rightarrow (a-h{)}^2+(0-k{)}^2=r^2$

$\Rightarrow (a+h{)}^2+(0-k{)}^2=r^2$

Subtract two equations

$\Rightarrow (a+h{)}^2=(a-h{)}^2$

$\Rightarrow 2ah=0$

$\Rightarrow h=0$

Substitute h=0 in above equation

$\Rightarrow r^2=a^2+k^2$

Therefore circle equation becomes :

$x^2+(y-k{)}^2=a^2+k^2$

Differentiate on both sides

$\Rightarrow 2x+2(y-k)y^{{}^{\prime }}=0$

$\Rightarrow k=\frac{x+y{y}^{{}^{\prime }}}{y^{{}^{\prime }}}=\frac{x}{y^{{}^{\prime }}}+y$

Substitute k in circle equation

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

$\Rightarrow x^2=a^2+y^2+2y.\frac{x}{y^{{}^{\prime }}}$

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