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(A) The given equation is $x^2+y^2+2x+2y-5=0$. Let the axes be rotated by an angle $\alpha$. The transformation equations are $x = X \cos \alpha - Y \

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(A) The given equation is $x^2+y^2+2x+2y-5=0$. Let the axes be rotated by an angle $\alpha$. The transformation equations are $x = X \cos \alpha - Y \sin \alpha$ and $y = X \sin \alpha + Y \cos \alpha$. Substituting these into the equation: $(X \cos \alpha - Y \sin \alpha)^2 + (X \sin \alpha + Y \cos \alpha)^2 + 2(X \cos \alpha - Y \sin \alpha) + 2(X \sin \alpha + Y \cos \alpha) - 5 = 0$. Simplifying the quadratic terms: $X^2(\cos^2 \alpha + \sin^2 \alpha) + Y^2(\sin^2 \alpha + \cos^2 \alpha) = X^2 + Y^2$. Simplifying the linear terms: $2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha)$. Thus,the transformed equation is $X^2 + Y^2 + 2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha) - 5 = 0$. For the equation to have no linear terms,the coefficients of $X$ and $Y$ must be zero: $\cos \alpha + \sin \alpha = 0 \implies 	an \alpha = -1$ $\cos \alpha - \sin \alpha = 0 \implies 	an \alpha = 1$ Since $	an \alpha$ cannot be both $1$ and $-1$ simultaneously,there is no value of $\alpha$ that satisfies both conditions. Therefore,the number of values of $\alpha$ is $0$.

If the axes are rotated through an angle $\alpha$,then the number of values of $\alpha$ such that the transformed equation of $x^2+y^2+2x+2y-5=0$ contains no linear terms is

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