TS EAMCET 4 May 2018 Shift 2 Solved Paper

The equation of the circle with radius a and center (a,a) isgiven by,
(x−a)2+(y−a)2=a2 .....(I)
x2+y2−2ax−2ay+a2=0
Differentiate equation (I) w.r.t x ,
2x+2yy′−2a−2ay′=0
x+yy′−a(1+y′)=0
a=

x+yy′

1+y′

Substitute value of a in equation (I).
(x−

x+yy′

1+y′

)+(y−

x+yy′

1+y′

) =(

x+yy′

1+y′

)2
(xy′−yy′)+(y−x)2=(x+yy′)2
(x−y)2(y′)2+(x−y)2=(x+yy′)2
(x−y)2[1+(y′)2]=(x+yy′)2