$C$ passes through the points $(2,4)$ and is such that the segment of any of its tangents at any point contained between the co-ordinate axis is biscected at the point of tangency. Let $S$ denotes circle described on the foci ${F_1}$ and ${F_2}$ of the conic $C$ as diameter.
Equation of the circle $S$ is

Let the point of tangency be $P = (x_0, y_0)$ on $C$.

Let tangent at $P$ meet the $x$-axis at $A$ and $y$-axis at $B$.

Given: $P$ is the midpoint of segment $AB$.

Let tangent line slope be $m$.

Equation of tangent:

$y - y_0 = m(x - x_0)$

Find $A$ and $B$:

• $A$: set $y = 0$:

$0 - y_0 = m(x_A - x_0) \implies x_A = x_0 - \frac{y_0}{m}$

So $A = (x_0 - \frac{y_0}{m}, 0)$.

• $B$: set $x = 0$:

$y_B - y_0 = m(0 - x_0) \implies y_B = y_0 - mx_0$

So $B = (0, y_0 - mx_0)$.

Midpoint condition: $P$ is midpoint of $A$ and $B$:

Midpoint coordinates:

$\left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right) = (x_0, y_0)$

So:

$\frac{\left(x_0 - \frac{y_0}{m}\right) + 0}{2} = x_0$

$x_0 - \frac{y_0}{m} = 2x_0$

$-\frac{y_0}{m} = x_0$

$m = -\frac{y_0}{x_0}$

Also from $y$-coordinate of midpoint:

$\frac{0 + (y_0 - mx_0)}{2} = y_0$

$y_0 - mx_0 = 2y_0$

$-mx_0 = y_0$

$m = -\frac{y_0}{x_0}$

Same equation — consistent.

So slope of tangent at $(x_0, y_0)$ is $-y_0/x_0$.

**Step 2: Find the curve $C$**

Slope $m = dy/dx = -y/x$.

Separate variables:

$\frac{dy}{dx} = -\frac{y}{x}$

$\frac{dy}{y} = -\frac{dx}{x}$

Integrate:

$\ln y = -\ln x + \ln k$

$\ln y = \ln \left(\frac{k}{x}\right)$

$y = \frac{k}{x}$

So $C$ is a rectangular hyperbola $xy = k.$

**Step 3: Use point (2, 4)**

$2 \cdot 4 = k \implies k = 8.$

Thus $C: xy = 8$, or $y = 8/x.$

**Step 4: Recognize conic type**

$xy = 8$ is a rectangular hyperbola.

Standard form: rotate coordinates by $45^\circ$.

Let $x = \frac{X+Y}{\sqrt{2}}$, $y = \frac{X-Y}{\sqrt{2}}$.

Then:

$xy = \frac{X^2 - Y^2}{2} = 8$

$X^2 - Y^2 = 16$

This is a rectangular hyperbola ($a=4$, $b=4$, $c=\sqrt{a^2 + b^2} = \sqrt{32} = 4\sqrt{2}$).

**Step 5: Foci in $XY$-coordinates**

For hyperbola $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$, foci at $(\pm c, 0)$ in $XY$-coordinates.

Here $a^2 = 16, b^2 = 16$, so $c^2 = 32, c = 4\sqrt{2}$.

So foci in $XY$-space: $F_1 = (4\sqrt{2}, 0), F_2 = (-4\sqrt{2}, 0)$.

**Step 6: Transform foci back to $xy$-coordinates**

$x = \frac{X+Y}{\sqrt{2}}, y = \frac{X-Y}{\sqrt{2}}$.

- $F_1$: $X = 4\sqrt{2}, Y = 0$

$x = \frac{4\sqrt{2}}{\sqrt{2}} = 4, y = \frac{4\sqrt{2}}{\sqrt{2}} = 4$.

So $F_1 = (4, 4)$.

- $F_2$: $X = -4\sqrt{2}, Y = 0$

$x = \frac{-4\sqrt{2}}{\sqrt{2}} = -4, y = \frac{-4\sqrt{2}}{\sqrt{2}} = -4$.

So $F_2 = (-4, -4)$.

Diameter endpoints: $(4, 4)$ and $(-4, -4)$.

Center = midpoint = $(0, 0)$.

Radius = distance from center to $F_1 = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$.

Equation of circle:

$x^2 + y^2 = (4\sqrt{2})^2 = 32$