Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$ 

Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$

 A: $O$ lies on the circle of diameter $AB$ and also on the circle of diameter $CD$, thus it must be one of their intersections (if any). But such circles are tangent, as the sum of their radii is equal to the distance $EF$ between their centers. It follows that $O$ is the common tangency point between those circles and lies therefore on $EF$.
By similar triangles (see diagram below), it is then immediate to compute $OG=3$ and $OH=6$. Hence $AO=\sqrt{OG^2+OH^2}=3\sqrt5$.

A: $A = (0,0)\\
B = (15,0)\\
C = (0,15)
D = (0,5)\\
O = (x,y)\\
(x,y)\cdot(x-15,y) = 0\\
x^2 + y^2 = 15x\\
(x,y-5)\cdot(x,y-15) = 0\\
x^2 + y^2 - 20y + 75 = 0\\
20y = 15 x + 75\\
y= \frac 34 x + \frac {15}{4}\\
x^2 + (\frac 34 x + \frac {15}{4})^2 = 15x\\
16x^2 + (3 x + 15)^2 = 240x\\
25x^2 + (90-240) x + 225 = 0\\
x^2 -6x + 9 = 0\\
(x-3)^2 = 0\\
x = 3\\
y = \frac 94 + \frac {15}{4} = 6\\
\|OA\| = \sqrt {3^2 + 6^2} = 3\sqrt 5$ 
If you have not learned about Euclidean inner products, we can do it with the Pythagorean theorem
$\|AO\|$ is the length of $AO$
$\|AO\| = \sqrt {x^2 + y^2}\\
\|OB\| = \sqrt {(15-x)^2 + y^2}\\
\|AO\|^2 + \|OB\|^2 = \|AB\|\\
x^2 + y^2 + x^2 - 30 x + 15^2 + y^2 = 15^2\\
x^2 + y^2 = 15 x$ 
and similarly 
$\|DO\|^2 + \|OC\|^2 = \|DC\|\\
x^2 + (y-5)^2 + x^2 + (y-15)^2  = 10^2\\
2x^2 + y^2 - 10y + 25 y^2-30 y + 225 = 100\\
2x^2 + 2y^2 - 40y + 150 = 0\\
x^2 + y^2 - 20y + 75 = 0$
A: 
\begin{align}
x=|OA|&=|CO_1|=|A_1O_2|=|BO_3|
.
\end{align}
Let $|AD|=a$, $|DC|=2a$. 
\begin{align}
|CO|&=2a\cos\theta
,\\
|AO|=|CO_1|&=
\sqrt{(3\,a)^2+(2a\,\cos\theta)^2-2\cdot3\,a\cdot(2\,a\,\cos\theta)
 \cdot\cos\theta}
\\
&=a\,\sqrt{9-8\,\cos^2\theta}
,\\
|AO_1|&=\sqrt{|AC|^2-|CO_1|^2}
\\
&=\sqrt{(3a)^2-a^2\,(9-8\cos^2\theta)}
\\
&=2\sqrt2\,a\,\cos\theta
,\\
|OO_1|&=
\sqrt{|CO|^2-|CO_1|^2}
\\
&=\sqrt{12\,a^2\cos^2\theta-a^2\,(9-8\,\cos^2\theta)}
\\
&=a\,\sqrt{4\cos^2\theta-9}
,\\
[ACA_1B]=(3a)^2&=
2\,|CO_1|\,|AO_1|+|OO_1|^2
\\
&=
2\,a\,\sqrt{9-8\,\cos^2\theta}
\cdot 2\sqrt2\,a\,\cos\theta
+\left(a\,\sqrt{12\cos^2\theta-9}\right)^2
.
\end{align}
The last equation simplifies to
\begin{align}
-18&+4\sqrt2\sqrt{9-8\cos^2\theta}\,\cos\theta
+12\,\cos^2\theta=0
,\\
(18-12\,\cos^2\theta)^2
&=
32\,(9-8\cos^2\theta)\,\cos^2\theta
,\\
4\,(10\cos^2\theta-9)^2&=0
,\\
\cos^2\theta&=\tfrac9{10},
\\
\text{and }\quad |AO|&=
a\,\sqrt{9-8\,\cos^2\theta}
=
5\,\sqrt{9-8\cdot\tfrac9{10}}
=3\,\sqrt5
.
\end{align}
