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P=y+x
P−Q=x+y−y'−(y')²
P=y−x
P+Q=1−x+y+y'+(y')²
We have ,(x−h)²+(y−h)²=r²
Differentiating and dividing by 2, x−h+yy'−hy'=0 ⇒h=(x+yy')/(1+y')
Again differentiating wrt x, hy''=1+yy''+(y')²
Putting value of h, ((x+yy')/(1+y'))y''=1+yy''+(y')²
(x+yy')y''=(1+yy''+(y')²)(1+y')
xy''+yy'y''=1+yy''+(y')²+y'+yy'y''+(y')³
xy''=1+yy''+(y')²+y'+(y')³
1+yy''+(y')²+y'³+yy''−xy''=0
1+y'(1+y'+(y')²)+y''(y−x)=0
⇒P=y−x,Q=1+y'+(y')²
⇒P+Q=1−x+y+y'+(y')²