Question

If the differential equation representing the family of all circles touching $x$-axis at the origin is $\left(x^2-y^2\right)\frac{dy}{dx}=g\left(x\right)y,$ then $g(x)$ equals :

• A. $\frac{1}{2}x$
• B. $2x^2$
• C. $2x$
• D. $\frac{1}{2}{x}^2$

Answer 1

Answer: (C)

$x^2+{\left(y-a\right)}^2=a^2$
$x^2+y^2-2ay=0...\left(i\right)$
diff. $w.r.t.x$
$2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0$
$a=\frac{x+y.y^{\prime }}{y^{\prime }}...\left(ii\right)$
put $\left(ii\right)$ in $\left(i\right)$
$x^2+y^2-2y\left(\frac{xy.y^{\prime }}{y^{\prime }}\right)=0$
$\left(x^2-y^2\right)y^{\prime }=2xy...\left(iii\right)$
compare $\left(iii\right)$ with $\left(x^2-y^2\right)\frac{dy}{dx}=g\left(x\right).y$
gives $g\left(x\right)=2x$