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 $OA=?$


$A = (0,0)\\
B = (15,0)\\
C = (0,15)
D = (0,5)\\
O = (x,y)\\
(x,y)\cdot(x-15,y) = 0\\
x^2 + y^2 = 15x\\
(x,y-5)\cdot(x,y-15) = 0\\
x^2 + y^2 - 20y + 75 = 0\\
20y = 15 x + 75\\
y= \frac 34 x + \frac {15}{4}\\
x^2 + (\frac 34 x + \frac {15}{4})^2 = 15x\\
16x^2 + (3 x + 15)^2 = 240x\\
25x^2 + (90-240) x + 225 = 0\\
x^2 -6x + 9 = 0\\
(x-3)^2 = 0\\
x = 3\\
y = \frac 94 + \frac {15}{4} = 6\\
\|OA\| = \sqrt {3^2 + 6^2} = 3\sqrt 5$ 
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 
$\|DO\|^2 + \|OC\|^2 = \|DC\|\\
x^2 + (y-5)^2 + x^2 + (y-15)^2  = 10^2\\
2x^2 + y^2 - 10y + 25 y^2-30 y + 225 = 100\\
2x^2 + 2y^2 - 40y + 150 = 0\\
x^2 + y^2 - 20y + 75 = 0$
