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

We need to find the area of a circle that passes through three given points: $(0,0)$, $(4,0)$, and $(3,9)$.

When a circle passes through three points, those points lie on the circumference of the circle. The center of such a circle is called the circumcenter.

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The circumcenter is equidistant from all three vertices. Let's call the circumcenter $(x, y)$.

Since all three points lie on the circle, they are all the same distance (the radius) from the center.

So we have:

Distance from center to $(0,0)$ = Distance from center to $(4,0)$

Distance from center to $(0,0)$ = Distance from center to $(3,9)$

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Using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Distance from $(x,y)$ to $(0,0)$ = Distance from $(x,y)$ to $(4,0)$

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

$x^2 + y^2 = (x-4)^2 + y^2$

Distance from $(x,y)$ to $(0,0)$ = Distance from $(x,y)$ to $(3,9)$

$\sqrt{x^2 + y^2} = \sqrt{(x-3)^2 + (y-9)^2}$

$x^2 + y^2 = (x-3)^2 + (y-9)^2$

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From the first equation:

$x^2 + y^2 = (x-4)^2 + y^2$

$x^2 + y^2 = x^2 - 8x + 16 + y^2$

$0 = -8x + 16$

$8x = 16$

$x = 2$

From the second equation:

$x^2 + y^2 = (x-3)^2 + (y-9)^2$

$x^2 + y^2 = x^2 - 6x + 9 + y^2 - 18y + 81$

$0 = -6x + 9 - 18y + 81$

$6x + 18y = 90$

Substituting $x = 2$:

$6(2) + 18y = 90$

$12 + 18y = 90$

$18y = 78$

$y = \tfrac{78}{18} = \tfrac{13}{3}$

Therefore, the circumcenter is $(2, \tfrac{13}{3})$

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The radius is the distance from the circumcenter to any of the three points. Using $(0,0)$:

$R = \sqrt{(2-0)^2 + (\tfrac{13}{3}-0)^2}$

$R = \sqrt{4 + \tfrac{169}{9}}$

$R = \sqrt{\tfrac{36}{9} + \tfrac{169}{9}}$

$R = \sqrt{\tfrac{205}{9}}$

Therefore: $R^2 = \tfrac{205}{9}$

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The area of a circle is $\pi R^2$

Area $= \pi \times \tfrac{205}{9} = \tfrac{205\pi}{9}$

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The area of the circle passing through the three given points is $\tfrac{205\pi}{9}$