Please Help

Ahh! Finally!!!! I think I found another way!!!!

Here's my own drawing...it's not as accurate as CPhill's :

Each of the purple lines is a radius of the circle,  " r ". Each radius meets the sides of the kite at right angles because each side of the kite is tangent to the circle.

Notice that there are two kites formed within the big kite. We can be sure that both these inside kites are similar to kite ABCD because all the corresponding angles are the same.

Looking at the kite ABCD to the little kite, we can say...

\(\frac{24}{18}\,=\,\frac{r}{18-r} \\~\\ 24(18-r)\,=\,18r \\~\\ 432-24r\,=\,18r \\~\\ 432\,=\,42r \\~\\ \frac{432}{42}\,=\,r \\~\\ r\,=\,\frac{72}{7}\qquad\text{cm}\)        Cross multiply

Here's one method  [ but maybe not the fastest or easiest ]

Let  AC   =  √ [ 24^2  + 18^2]  =  √ 900  =  30

So....let  A  = (-15, 0)  and C  =  (15, 0)

And we can find B and D  by intersecting the circles

(x + 15)^2 +  y^2  = 24^2      and

(x - 15)^2  + y^2  = 18^2

So....the intersection of  these (arbitrarily)  gives  B = (4.2, 14.4)  and D  = (4.2, -14.4) 

Here's a pic :

Let the center of the circle   lie at   ( E, 0)  [ I'm using E for the center instead of O ]

The slope of AB  is  [ 14.4] / [ 4.2 - - 15]  =  14.4  / 19.2  =  3/4

And the equation of this line segment is 

y =  (3/4)(x -  -15)

y =  (3/4)x + 45/4

4y = 3x + 45

3x - 4y + 45  =  0

And the slope  of BC  is  -4/3

And the equation of this line segment  is

y  =  (-4/3)(x - 15)

y =  (-4/3)x + 20

3y  = -4x + 60

4x + 3y - 60  =  0

And the distance   between  a point (m, n)  and the line  Ax + By + C = 0  is given by

l  Am + Bn  + C l   /  √ [ A^2 + B^2  ]

So  we want to  equate  these and solve for  E

l  3E  - 4 (0)  +  45 l  / √ [ 3^2 + (-4)^2]   =  l 4E + 3(0)  - 60 l / √[4^2 + 3^2 ]

l  3E +  45 l  /  5   =  l 4E - 60 l / 5

We have these two possible equations

3E + 45   =  4E - 60

E  =   105        reject

Or

3E + 45  =  -  [ 4E - 60 ]

3E + 45  =  -4E + 60

7E  =  15

E =  15/7

So....the center  of  of the circle  is  (E, 0)  =  (15/7, 0)

And using   l 3C + 45 l  /  5....the radius of the circle is

l 3 (15/7) + 45 l  /  5  =    l 45/7  +  45 l  / 5  =   45/35 + 9  =  9/7 + 9  =  72/7 cm

So.....the equation of the inscribed circle is

(x - 15/7)^2  + y^2  =  (72/7)^2

Ah!!!....that's more elegant than mine, for sure  !!!