TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Given circle
x2+y2−4x−2y−4=0
Centre (2,1) Equation of diameter of circle passes through origin.
x−2y=0
Equation of circle through the end point of diameter of circle and line x−2y=0 is
x2+y2−4x−2y−4+λ(x−2y)=0
Since, this is passes through (1,2)
∵ 1+4−4−4−4+λ(−3)‌‌=0
λ‌‌=‌

−7

3

Required equation of circle is
x2+y2−4x−2y−4−‌

7

3

(x−2y)=0
3(x2+y2)−19x+8y−12=0
3x2+3y2−19x+8y−12=0