The general equation of a circle in first quadrnt and touching the coordinate axes is given by
`(x-a)^(2)+(y-a)^(2)=a^(2)" "...(i)"where a is parameter."`
Since this equation contains one parameter, so we will differentiate it only once to get the differential equation. On differentiating (i) w.r.t,x, we get
`2(x-a)+2(y-a)(dy)/(dx)=0`
`implies(x-a)+ (y-a)p=0where (dy)/(dx)=0`
`implies(x-a)+(y-a)=-0, where (dy)/(dx)=p`
`impliesz+yp=a(1+p)a=((x+yp))/((1+p))." "...(ii)`
Putting the bvalue of a from (ii) in (i), we get
`(x-(x+yp)/(1+p))^(2)+(y-(x+yp)/(1+p))^(2)=((x+yp)/(1+p))^(2)`
`implies(xp-yp)^(2)+(y-x)^(2)=(x+yp)^(2)`
`implies(x-y)^(2)p^(2)+(x-y)^(2)=(x+y)^(2)`
`implies(x-y)^(2)(p^(2)+1)=(x+yp)^(2)`
`implies(x-y)^(2){((dy)/(dx))^(2)+1}=(x+y(dy)/(dx))^(2),`
which is the required differential equation.
