If the axes are rotated through an angle alph | vedclass

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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,we get $X^2 + Y^2 + 2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha) - 5 = 0$. For the equation to contain no linear terms,the coefficients of $X$ and $Y$ must be zero: $1) \cos \alpha + \sin \alpha = 0$ $\Rightarrow 	an \alpha = -1$ $\Rightarrow \alpha = n\pi - \frac{\pi}{4}$. $2) \cos \alpha - \sin \alpha = 0$ $\Rightarrow 	an \alpha = 1$ $\Rightarrow \alpha = n\pi + \frac{\pi}{4}$. Since there is no value of $\alpha$ that satisfies both conditions simultaneously,the number of such values 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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