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(C) The given curves are $x^{2}+y^{2}=4x$ and $x^{2}+y^{2}=8$.For the first curve $x^{2}+y^{2}=4x$,differentiating with respect to $x$ gives $2x + 2y

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$\frac{1}{\sqrt{2}}$

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(C) The given curves are $x^{2}+y^{2}=4x$ and $x^{2}+y^{2}=8$.For the first curve $x^{2}+y^{2}=4x$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2-x}{y}$.At the point $(2,2)$,the slope $m_{1} = \frac{2-2}{2} = 0$.For the second curve $x^{2}+y^{2}=8$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$.At the point $(2,2)$,the slope $m_{2} = -\frac{2}{2} = -1$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} ight|$.Substituting the values,$	an 	heta = \left| \frac{0 - (-1)}{1 + (0)(-1)} ight| = \left| \frac{1}{1} ight| = 1$.Since $	an 	heta = 1$,we have $	heta = 45^{\circ}$.Given that $	heta = \sin^{-1} a$,we have $\sin^{-1} a = 45^{\circ}$,which implies $a = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.

If $\sin ^{-1} a$ is the acute angle between the curves $x^{2}+y^{2}=4 x$ and $x^{2}+y^{2}=8$ at $(2,2)$,then $a$ is equal to

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