Question
Without changing the direction of coordinate axes, origin is transferred to (h, k), so that the linear (one degree)
terms in the equation x2+y2−4x+6y−7=0 are eliminated. Then the point (h, k) is
terms in the equation x2+y2−4x+6y−7=0 are eliminated. Then the point (h, k) is
Answer: Option C
:
C
Putting x = x' + h, y = y' + k, the given equation transforms to
x′2+y′2+x′(2h−4)+y′(2k+6)+h2+k2−7=0
To eliminate linear terms, we should have
2h - 4 = 0, 2k + 6 = 0 ⇒ h = 2, k = -3
i.e., (h, k) = (2, -3).
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:
C
Putting x = x' + h, y = y' + k, the given equation transforms to
x′2+y′2+x′(2h−4)+y′(2k+6)+h2+k2−7=0
To eliminate linear terms, we should have
2h - 4 = 0, 2k + 6 = 0 ⇒ h = 2, k = -3
i.e., (h, k) = (2, -3).
Was this answer helpful ?
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