AP EAPCET 25-Aug-2021 Shift 2 Paper

Equation of diameters of the circle,
x+2y−5=0 ...(i)
3x−y−1=0 ...(ii)
Since, intersection of lines (i) and (ii) is the centre of the circle.
On solving Eqs. (i) and (ii), we get x=1 and y=2
∴ Centre =(1,2) and radius =5
∴ Equation of required circle
(x−1)2+(y−2)2=(5)2
⇒ x2+y2−2x−4y−20=0