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(C) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Let $P(5 \cos 	heta, 4 \sin 	heta)$ be a point on the ellipse.The equation

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$(0, 0)$

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(C) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Let $P(5 \cos 	heta, 4 \sin 	heta)$ be a point on the ellipse.The equation of the tangent at $P$ is $\frac{x(5 \cos 	heta)}{25} + \frac{y(4 \sin 	heta)}{16} = 1$,which simplifies to $\frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{4} = 1$.This tangent intersects the $X$-axis at point $A$ by setting $y=0$,giving $A = (\frac{5}{\cos 	heta}, 0) = (5 \sec 	heta, 0)$.The image of $A(5 \sec 	heta, 0)$ with respect to the line $y=x$ is $A^{\prime} = (0, 5 \sec 	heta)$.The equation of the circle with diameter $AA^{\prime}$ is $(x - 5 \sec 	heta)(x - 0) + (y - 0)(y - 5 \sec 	heta) = 0$.This simplifies to $x^2 + y^2 - 5 \sec 	heta (x + y) = 0$.For this circle to pass through a fixed point independent of $	heta$,we observe that for any $	heta$,the point $(0, 0)$ satisfies the equation $0^2 + 0^2 - 5 \sec 	heta (0 + 0) = 0$.

$A$ tangent drawn at a point on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ cuts the $X$-axis at point $A$. If $A^{\prime}$ is the image of $A$ with respect to the line $y=x$,then the circle with $AA^{\prime}$ as its diameter passes through the fixed point:

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