Question

Obtain the differential equation of the family of circles having centre at (0, b) and passing through the points (a, 0) and (-a, 0), where 'b' is the arbitrary constant.

Answer 1

As $(x-h{)}^2+(y-k{)}^2=r^2$ is the circle with centre at (h, k) and radius of r.

So, $(x-0{)}^2+(y-b{)}^2=r^2,$ where $r^2=(a-0{)}^2+(0-b{)}^2=a^2+b^2$

$\Rightarrow x^2+y^2-2by+b^2=a^2+b^2\Rightarrow x^2+y^2-2by=a^2...\left(i\right)$

$\therefore 2x+2y\frac{dy}{dx}-2b\frac{dy}{dx}=0\Rightarrow b=\frac{x+y{y}^{\prime }}{y^{\prime }}$

Substituting value of b in (i), $x^2+y^2-2\left(\frac{x+y{y}^{\prime }}{y^{\prime }}\right)y=a^2$

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

That is, $(x^2-y^2-a^2)\frac{dy}{dx}-2xy=0$ is the required differential equation.