Without changing the direction of coordinates axes, origin is transferred to $$($$α,β$$)so$$ that the linear terms in the equation $$x2$$ $$+$$ $$y2$$ $$+$$ $$2x$$ – $$4y$$ $$+$$ $$6$$ $$=$$ $$0$$ are eliminated. The point $$($$α,β$$)$$ is

Question

Without changing the direction of coordinates axes, origin is transferred to $$($$α,β$$)so$$ that the linear terms in the equation $$x2$$ $$+$$ $$y2$$ $$+$$ $$2x$$ – $$4y$$ $$+$$ $$6$$ $$=$$ $$0$$ are eliminated. The point $$($$α,β$$)$$ is

Infinity Learn · Accepted Answer

The given equation $$isx2$$ $$+$$ $$y2$$ $$+$$ $$2x$$ – $$4y$$ $$+$$ $$6$$ $$=$$ $$0$$                                          (1)Putting$$ x $$=$$ x′ $$+$$ α, y $$=$$ y′ $$+$$ β in $$(1), we getx′$$2$$ $$+$$ y′$$2$$ $$+$$ x′(2$$α$$+$$ $$2) $$+$$ y′(2$$β – $$4) $$+$$ $$($$α$$2$$ $$+$$ β $$2$$ $$+$$ $$2$$α – $$4$$β$$+$$ $$6)$$ $$=$$ $$0To$$ eliminate linear terms, we should have $$2$$α $$+$$ $$2$$ $$=$$ $$0$$ and $$2$$β– $$4$$ $$=$$ $$0$$⇒ α $$=$$ –$$1$$ and β $$=$$ $$2$$∴$$($$α, β$$)$$ ≡ $$($$–$$1$$, $$2)$$

Without changing the direction of coordinates axes, origin is transferred to $(α, β)$ so that the linear terms in the equation x2 + y2 + 2x – 4y + 6 = 0 are eliminated. The point $(α, β)$ is

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Without changing the direction of coordinates axes, origin is transferred to (α,β)so that the linear terms in the equation x2 + y2 + 2x – 4y + 6 = 0 are eliminated. The point (α,β) is

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Without changing the direction of coordinates axes, origin is transferred to $(α, β)$ so that the linear terms in the equation x2 + y2 + 2x – 4y + 6 = 0 are eliminated. The point $(α, β)$ is