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

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$(0,0)$

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(C) The equation of the given ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Let a point $P(5 \cos 	heta, 4 \sin 	heta)$ be on the ellipse. The equation of the tangent to the ellipse at point $P$ is $\frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{4} = 1$. Setting $y=0$ to find the $X$-axis intercept $Q$,we get $x = \frac{5}{\cos 	heta}$. Thus,$Q = (5 \sec 	heta, 0)$. Since $R$ is the image of $Q$ with respect to $y=x$,we swap the coordinates to get $R = (0, 5 \sec 	heta)$. The equation of a circle with diameter $QR$ is $(x - x_Q)(x - x_R) + (y - y_Q)(y - y_R) = 0$. Substituting the coordinates,we get $(x - 5 \sec 	heta)(x - 0) + (y - 0)(y - 5 \sec 	heta) = 0$. This simplifies to $x^2 + y^2 - (5 \sec 	heta)x - (5 \sec 	heta)y = 0$. For any value of $	heta$,the point $(0,0)$ satisfies this equation. Therefore,the circle always passes through the origin $(0,0)$.

Let a tangent drawn at any point on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut the $X$-axis at $Q$. Let $R$ be the image of $Q$ with respect to $y=x$. If $S$ is a circle with $QR$ as its diameter,then the fixed point through which the circle $S$ passes is

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