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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 $A^{\prime}$ be the image of $A$ with respect to the line $y=x$, then the circle with $A A^{\prime}$ as its diameter passes through the fixed point
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$(0,0)$
Given equation of ellipse
$\frac{x^2}{25}+\frac{y^2}{16}=1$
Let $P(5 \cos \theta, 4 \sin \theta)$ be a point on $16 x^2+25 y^2=400$.
The equation of the tangent at $P$ is
$4 x \cos \theta+5 y \sin \theta=20$
This meets the coordinate axes at $A(5 \sec \theta, 0)$ and $A^{\prime}(0,4 \operatorname{cosec} \theta)$
The equation of the circle with $A A^{\prime}$ as diameter is
$\begin{aligned} & (x-5 \sec \theta)(x-0)+(y-0)(y-4 \operatorname{cosec} \theta)=0 \\ & \Rightarrow x^2+y^2-5 x \sec \theta-4 y \operatorname{cosec} \theta=0\end{aligned}$
Clearly, it passes through $(0,0)$
$\frac{x^2}{25}+\frac{y^2}{16}=1$
Let $P(5 \cos \theta, 4 \sin \theta)$ be a point on $16 x^2+25 y^2=400$.
The equation of the tangent at $P$ is
$4 x \cos \theta+5 y \sin \theta=20$
This meets the coordinate axes at $A(5 \sec \theta, 0)$ and $A^{\prime}(0,4 \operatorname{cosec} \theta)$
The equation of the circle with $A A^{\prime}$ as diameter is
$\begin{aligned} & (x-5 \sec \theta)(x-0)+(y-0)(y-4 \operatorname{cosec} \theta)=0 \\ & \Rightarrow x^2+y^2-5 x \sec \theta-4 y \operatorname{cosec} \theta=0\end{aligned}$
Clearly, it passes through $(0,0)$
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