AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

Given equation, is y2−6y−4x+13=0
Let origin to be shifted to the point (h,k)
So, put x→x+h
y→y+k in Eqq(i)
From To eliminate y and constant terms
put coefficient of y=0 and constant =0
2k−6=0‌‌ and ‌‌k2−6k−4h+13=0
k=3‌‌9−6(3)−4h+13=0
−9−4h+13=0
h=1
∴ Required point =(h,k)=(1,3)