Let a tangent drawn at any point on the ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{16}=1$ 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 it’s  diameter, then the fixed point through which the circle S  passes is

Given ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{16}=1 (\mathrm{a}>\mathrm{b})$

where ${\mathrm{a}}^2=25, {\mathrm{b}}^2=16$

Let $\mathrm{P}=(5 \mathrm{cos\theta }, 4 \mathrm{sin\theta })$ be any point on ellipse

Now equation of tangent at $\mathrm{P}\left(\mathrm{\theta }\right)$ is $\frac{\mathrm{x}}{5} \mathrm{cos\theta }+\frac{\mathrm{y}}{4}\mathrm{sin\theta }=1$

It cuts x-axis at $\mathrm{Q}=\left(\frac{5}{\mathrm{cos\theta }},0\right)$

Given R=image of Q w.r to y=x

$\Rightarrow \mathrm{R}=\left(0,\frac{5}{\mathrm{cos\theta }}\right)$

Now equation of circle having $\overline{\mathrm{QR}}$ as diameter is$\left(\mathrm{x}-{\mathrm{x}}_1\right)(\mathrm{x}-{\mathrm{x}}_2)+(\mathrm{y}-{\mathrm{y}}_1)(\mathrm{y}-{\mathrm{y}}_2)=0$

$$\begin{gathered} \Rightarrow \left(\mathrm{x}-\frac{5}{\mathrm{cos\theta }}\right)\mathrm{x}+\left(\mathrm{y}\right)\left(\mathrm{y}-\frac{5}{\mathrm{cos\theta }}\right)=0 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-\frac{5\mathrm{x}}{\mathrm{cos\theta }}-\frac{5\mathrm{y}}{\mathrm{cos\theta }}=0 \end{gathered}$$

It passes through the point (0, 0)