) 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