(a) Given equation of circle,
(x−h)2+(y−k)2=1...(i)On differentiating w.r.t x, we get,
⇒ 2(x−h)+2(y−k)=0⇒(x−h)+(y−k)=0...(ii)Again, diffierentiating w.r.t x, we get
⇒1+(y−k)+.=0⇒(y−k)+()2=−1...(iii)From Eqs. (ii) and (iii), we obtain
(y−K)=−and
(x−h)= On substituting these values in the Eq. (i), we get
[1+()2]3=()2which is the required differential equation.