Answer:
Option B,C
Explanation:
Since the centre lies on y=x.
$\therefore$ Equation of circle is
$x^{2}+y^{2}-2ax-2ay+c=0$
On differentiating , we get
$2x+2yy'-2a-2ay'=0$
$\Rightarrow$ $x+yy'-a-ay'=0$
$\Rightarrow$ $a=\frac{x+yy'}{1+y'}$
Again differentiating , we get
0 $= \frac{(1+y')[1+yy'+(y')^{2}]-(x+yy').(y")}{(1+y')^{2}}$
$\Rightarrow$ $ (1+y')[1+(y')^{2}+yy"]- (x+yy')(y")=0$
$\Rightarrow$ $1+y'[(y')^{2}+y'+1]+y"(y-x)=0$
On comparing with $Py"+Qy'+1=0,$ we get
P=y-x and $Q= (y')^{2}+y'+1$