Circle the Drain

This question was very long (at least by my method).

I used coordinate geometry.

After lots of calculations I arrived at the following conclusion. (See image below)

In the given image $I$ denotes the incenter , $C$ denotes the cicumcenter and $N$ denotes the center of nine point circle.

Now we could have just figured out the area by area formula.

$\text{ABC is the triangle with D as the foot of altitude from A. HNGO is a section of Euler Line in}\\ \text{blue. H - Orthocenter, N - Nine point center, G - Centroid, O - Circumcenter, I - Incenter.}\\ \text{We use coordinate geometry and find coordinates of N, I, O.}\\ \text{Coordinate of I is found as .} ~~~~ X_I=\dfrac{aX_A+bX_B+cX_C} s, ~~~~~~~~~~~Y_I=\dfrac{aY_A+bY_B+cY_C} s\\ \text{To simplify calculations, coordinate axis is chosen taking D as origin, and BC along X-axis.}\\ \text{So A(0,8), B(-6,0), C(15,0), a, b, c are sides of ABC. From Pythagoras, BD=6, DC=15.}\\ 2*s=a+b+c=21+17+10, \therefore ~s=24.\\ 24*X_I=21*0+17*(-6)+15*10=48, ~~24*Y_I=21*8+17*0+15*0=168. \ \ \ \color{#3D99F6}{I(2,7)}.\\ \text{In Euler Line, if we know two points, we can find out other points. It takes less calculations to}\\ \text{find G and H, since we know coordinates of A, B, C, one altitude AD, can easily find slope of CH.}\\\implies ~X_G=\dfrac{0-6+15}3=3, ~~~ ~~Y_G=\dfrac{8+0+0} 3=\dfrac 8 3. \ \ \ \ \ \ \ \ \ \ \ \ \ ~~~~~~~~\color{#3D99F6}{G(3,\frac 8 3)}.\\ \text{CE is the altitude from C, to meet BA extended at E, extended further to H to meet DA extended}.\\ Y_{BA}=\dfrac 8 6 X+8. ~~~~ CE \bot BA, \implies ~Y_{CE}-0=\dfrac{-6} 8(X-15), \implies ~ Y_{CE}=\dfrac{-3} 4*X -\dfrac{45} 4. \\ \text{So H on X=0, is } \color{#3D99F6}{ H(0, \frac{45} 4).} \\ \text{Properties of Euler Line.}\\ OH=\frac 3 2 *GH=\frac 3 2*\{G - H\}=\frac 3 2*\{ (3,\frac 8 3)-(0,\frac{45} 4) \}=(\frac 9 2,4-\frac{45*3}{4*2} )\\ \therefore ~H+OH=(\frac 9 2 ,4-\frac{45*3}{4*2}+\frac{45} 4)=(\frac 9 2 ,4-\frac{45}{4*2}\\ \text{Will come soon to complete}.$