If the origin is shifted to (2,3) and the axe | vedclass

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(B) The given equation is $2 x^2+2 y^2-8 x-12 y+18=0$.We can rewrite this by completing the square:$2(x^2-4x+4) + 2(y^2-6y+9) = -18 + 8 + 18$$2(x-2)^2

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$x^2+y^2=4$

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(B) The given equation is $2 x^2+2 y^2-8 x-12 y+18=0$.We can rewrite this by completing the square:$2(x^2-4x+4) + 2(y^2-6y+9) = -18 + 8 + 18$$2(x-2)^2 + 2(y-3)^2 = 8$$(x-2)^2 + (y-3)^2 = 4$.When the origin is shifted to $(2,3)$,the equation becomes $X^2 + Y^2 = 4$,where $X = x-2$ and $Y = y-3$.Rotating the axes by an angle $	heta = 45^{\circ}$ does not change the form of the circle $X^2 + Y^2 = r^2$,as $X^2 + Y^2 = (X' \cos 	heta - Y' \sin 	heta)^2 + (X' \sin 	heta + Y' \cos 	heta)^2 = X'^2 + Y'^2$.Thus,the transformed equation remains $x^2+y^2=4$.

If the origin is shifted to $(2,3)$ and the axes are rotated through an angle $45^{\circ}$ about that point,then the transformed equation of $2 x^2+2 y^2-8 x-12 y+18=0$ is

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