Consider the family of all circles whose centres lie on the straight line $y=x$. If this family of circles is represented by the differential equation $P y''+Q y'+1=0$, where $P, Q$ are functions of $x, y$ and $y'\left(\right.$ here $\left.y'=\frac{d y}{d x}, y''=\frac{d^2 y}{d x^2}\right)$, then which of the following statements is (are) true?

(a) $P=y+x$
(b) $P=y-x$
(c) $P+Q=1-x+y+y'+\left(y'\right)^2$
(d) $P-Q=x+y-y'-\left(y'\right)^2$

(b), (c)

The equation of the family of circles whose centres lie on the straight line $y=x$ is

$x^2+y^2-2 a x-2 a y+c=0$, whose a and c are parameters.

Differentiating with respect to $x$, we obtain

$2 x+2 y y'-2 a-2 a y'=0 \Rightarrow a=\frac{x+y y'}{1+y'}$

Differentiating this with respect to x, we obtain

$\frac{\left(1+y'\right)\left(1+\left(y'\right)^2+y y''\right)-\left(x+y y'\right) y''}{\left(1+y'\right)^2}=0$

$\Rightarrow \left(1+y'\right)\left(1+\left(y'\right)^2+y y''\right)-\left(x+y y'\right) y''=0$

$\Rightarrow 1+y'\left(\left(y'\right)^2+y'+1\right)+y''(y-x)=0$

Comparing this equation with $P y''+Q y'+1=0$, we obtain

$P =y-x$ and $Q=\left(y'\right)^2+y'+1$

$\Rightarrow P =y-x$ and $P+Q=1-x+y+y'+\left(y'\right)^2$

Hence, options (b) and (c) are correct.