AP EAPCET 19-Aug-2021 Shift 2 Paper

Equation of normal passing through (4,3) and (2,1)
y−3=

1−3

2−4

(x−4)
y−3=x−4
x−y=1 ...(i)
Equation of diameter
2x−y=2 ...(ii)
Solving Eqs. (i) and (ii), we have coordinate of centre of circle
∴ x=1,y=0
Centre of Circle (1,0)
Radius of circle = distance between (1,0) and (2,1)
=√(2−1)2+(1−0)2=√2
Radius =√2
∴ Equation of circle
(x−1)2+(y−0)2=(√2)2
⇒x2+y2−2x−1=0