The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is

The correct answer is $\frac{205\pi }{9}$.

We need to find the area of the circumscribed circle (circumcircle) passing through the three vertices of the triangle: A(0, 0), B(4, 0), and C(3, 9). The strategy is to find the circumradius R using the formula:

$R=\frac{abc}{4\times \text{Area of triangle}}$

Step 1: Find the side lengths of the triangle.

Side a = BC (from B(4,0) to C(3,9)):
$a=\sqrt{(4-3{)}^2+(0-9{)}^2}=\sqrt{1+81}=\sqrt{82}$

Side b = AC (from A(0,0) to C(3,9)):
$b=\sqrt{3^2+9^2}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}$

Side c = AB (from A(0,0) to B(4,0)):
$c=\sqrt{4^2+0^2}=4$

Step 2: Find the area of the triangle using the coordinate formula.

$\text{Area}=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Substituting A(0,0), B(4,0), C(3,9):
$\text{Area}=\frac{1}{2}\left|0\right(0-9)+4(9-0)+3(0-0\left)\right|$

$=\frac{1}{2}|0+36+0|=\frac{36}{2}=18$

Step 3: Calculate the circumradius R.

$R=\frac{abc}{4\times \text{Area}}=\frac{\sqrt{82}\times 3\sqrt{10}\times 4}{4\times 18}$

$=\frac{12\sqrt{82\times 10}}{72}=\frac{\sqrt{820}}{6}$

Now simplify $\sqrt{820}$:
$820=4\times 205$, so $\sqrt{820}=2\sqrt{205}$

$R=\frac{2\sqrt{205}}{6}=\frac{\sqrt{205}}{3}$

Step 4: Find the area of the circumcircle.

$R^2=\frac{205}{9}$

$\text{Area of circle}=\pi R^2=\frac{205\pi }{9}$

Therefore, the area of the circle passing through all three vertices of the triangle is $\frac{205\pi }{9}$.