1. Given parabola and its focus

The parabola is $y^2 = 8x.$ Comparing with the standard form $y^2 = 4ax$, we get $4a = 8 \implies a = 2.$ So the focus is $S = (2,0).$

2. Given circle

The circle is $x^2 + y^2 - 2x - 4y = 0.$

3. Find the common points of the parabola and the circle

From the parabola, $x = \frac{y^2}{8}.$ Substitute into the circle: $\left(\frac{y^2}{8}\right)^2 + y^2 - 2\left(\frac{y^2}{8}\right) - 4y = 0.$ That is, $\frac{y^4}{64} + y^2 - \frac{y^2}{4} - 4y = 0.$ So, $\frac{y^4}{64} + \frac{3y^2}{4} - 4y = 0.$ Multiply by $64$: $y^4 + 48y^2 - 256y = 0.$ Factor out $y$: $y(y^3 + 48y - 256) = 0.$ Now check simple roots of $y^3 + 48y - 256 = 0$. For $y=4$, $4^3 + 48(4) - 256 = 64 + 192 - 256 = 0.$ So, $y^3 + 48y - 256 = (y-4)(y^2+4y+64).$ Since $y^2+4y+64$ has no real roots, the real values are $y=0,\quad y=4.$

Thus the intersection points are:

• For $y=0$: $x=\frac{0}{8}=0 \implies P=(0,0).$
• For $y=4$: $x=\frac{16}{8}=2 \implies Q=(2,4).$

4. Area of triangle $PQS$

We have $P=(0,0), \quad Q=(2,4), \quad S=(2,0).$

Take $PS$ as the base. Since $P=(0,0)$ and $S=(2,0)$, $PS = 2.$ The perpendicular distance of $Q=(2,4)$ from the line $PS$ (which is the $x$-axis) is $4.$ So the area is $\text{Area} = \frac{1}{2} \times 2 \times 4 = 4.$

5. Comparison with stored answer

Derived answer is $4,$ which matches the stored correct answer.