TS EAMCET 7 May 2018 Shift 1 Solved Paper

Polar of point A(7,

−1

2

) with respect to circle x2+y2−4x+6y−12=0
7x−

1

2

y−2(x+7) +3(y−

1

2

)−12=0
14x−y−4x−28 +6y−3−24=0
10x+5y−55=0
2x+y−11=0
Normal to the circle at (1,3) is
x+2y=7
Normal to the circle at (3,5) is
2x+y=11
The intersection point of two normal is
x=5 and y=1.
Then equation of the circle with centre at (5,1) is
(x−5)2+(y−1)2 ={√(5−1)2+(1−3)2}2
x2+y2−10x−2y+6=0