Coordinate Geometry Question 13
Question: Without changing the direction of coordinate axes, origin is transferred to $ (h,k) $ , so that the linear (one degree) terms in the equation $ x^{2}+y^{2}-4x+6y-7 $ =0 are eliminated. Then the point $ (h,k) $ is
Options:
A) (3, 2)
B) (- 3, 2)
C) (2, - 3)
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ x={x}’+h,y={y}’+k, $ the given equation transforms to $ {{{x}’}^{2}}+{{{y}’}^{2}}+{x}’(2h-4)+{y}’(2k+6)+h^{2}+k^{2}-7=0 $
To eliminate linear terms, we should have $ 2h-4=0,2k+6=0\Rightarrow h=2,k=-3 $ i.e., $ (h,k)=(2,-3) $ .
BETA
