Q.Find the differential equation of the circles passing through the points of intersection of unit circle with centre at the origin and line bisecting the first quadrant.

The circles through the intersection of $x^2+y^2=1$ and $y=x$ form the one-parameter family $x^2+y^2-1+\lambda(y-x)=0$; differentiating once and eliminating $\lambda$ gives $\dfrac{dy}{dx}=\dfrac{1+x^2-y^2-2xy}{1-x^2+y^2-2xy}$.

Concept. All circles passing through the common points of a given circle $S=0$ and a line $L=0$ are given by $S+\lambda L=0$ for a parameter $\lambda$. One arbitrary constant means the differential equation is first order — formed by one differentiation and elimination of $\lambda$, a routine NCERT Class 12 mathematics procedure.

Form the family. Unit circle $x^2+y^2-1=0$; bisector of the first quadrant $y=x$, i.e. $y-x=0$. So

$x^2+y^2-1+\lambda(y-x)=0.\qquad(1)$

Differentiate w.r.t. $x$ (write $y'=\frac{dy}{dx}$):

$2x+2yy'+\lambda(y'-1)=0\ \Rightarrow\ \lambda=\frac{2x+2yy'}{1-y'}.\qquad(2)$

Also solve (1) for $\lambda$: $\lambda=\dfrac{1-x^2-y^2}{\,y-x\,}$.

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