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

• A. (3, 2)
• B. (- 3, 2)
• C. (2, - 3)
• D. None of these

Answer 1

The correct option is C

(2, - 3)

Putting x = x' + h, y = y' + k, the given equation transforms to

$x^{\prime 2}+y^{\prime 2}+x^{\prime }(2h-4)+y^{\prime }(2k+6)+h^2+k^2-7=0$

To eliminate linear terms, we should have

2h - 4 = 0, 2k + 6 = 0 $\Rightarrow$ h = 2, k = -3

i.e., (h, k) = (2, -3).