#### Question

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

#### Solution

We know that (x−a)^{2}+(y−b)^{2}=r^{2} represents a circle with centre (*a*, *b*) and radius *r*.â€‹

Since the circle lies in the 2nd quadrant, and touches the coordinate axes, thus *a *< 0, *b* > 0 and |*a*| = |*b*| = *r.*

So, the equation becomes (x+a)^{2}+(y−a)^{2}=a^{2} .....(1)

Differentiating this equation w.r.t. *x*, we get

`2(x+a)+2(y−a)dy/dx=0`

`⇒dy/dx=(−x+a)/(y−a)`

Putting `dy/dx=y',` we get

`y'=(−x+a)/(y−a)`

`⇒yy'−ay'+x+a=0`

`⇒yy'+x=ay'−a`

`⇒a=(x+yy')/(y'−1)`

Substituting this value of *a* in (1), we get

`(x−(x+yy')/(y'−1))^2+(y−(x+yy')/(y'−1))^2=((x+yy')/(y'−1))^2`

`⇒(xy'−x−x−yy')^2+(yy'−y−x−yy')^2=(x+yy')^2`

`⇒[y'(x−y)−2x]^2+(x+y)^2=(x+yy')^2`

`⇒(y')^2(x^2−2xy+y^2)−4x^2y'+4xyy'+4x^2+x^2+2xy+y^2=x^2+2xyy'+y^2(y')^2`

`⇒(y')^2(x^2−2xy)+2xy'(−2x+y)+4x^2+2xy+y^2=0`

This is the required â€‹differential equation of the family of circles in the second quadrant and touching the coordinate axes.