Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$
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$O$ 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 $AO=\sqrt{OG^2+OH^2}=3\sqrt5$.
Intelligenti pauca
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Because $O$ lies in the intersection between the circles of diameters $AB$ and $CD$, but such circles are tangent as the sum of their radii is equal to $EF$. It follows that $O$ is the common tangency point between those circles and lies therefore on $EF$. – Intelligenti pauca Dec 05 '17 at 10:15
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I explained that in my edited answer: $$EF=\sqrt{AF^2+AE^2}=\sqrt{7.5^2+10^2}=12.5,$$ thus the distance between the circle centers is equal to the sum of their radii. – Intelligenti pauca Dec 05 '17 at 10:32
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$A = (0,0)\\ B = (15,0)\\ C = (0,15) D = (0,5)\\ O = (x,y)\\ (x,y)\cdot(x-15,y) = 0\\ x^2 + y^2 = 15x\\ (x,y-5)\cdot(x,y-15) = 0\\ x^2 + y^2 - 20y + 75 = 0\\ 20y = 15 x + 75\\ y= \frac 34 x + \frac {15}{4}\\ x^2 + (\frac 34 x + \frac {15}{4})^2 = 15x\\ 16x^2 + (3 x + 15)^2 = 240x\\ 25x^2 + (90-240) x + 225 = 0\\ x^2 -6x + 9 = 0\\ (x-3)^2 = 0\\ x = 3\\ y = \frac 94 + \frac {15}{4} = 6\\ \|OA\| = \sqrt {3^2 + 6^2} = 3\sqrt 5$
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
$\|DO\|^2 + \|OC\|^2 = \|DC\|\\ x^2 + (y-5)^2 + x^2 + (y-15)^2 = 10^2\\ 2x^2 + y^2 - 10y + 25 y^2-30 y + 225 = 100\\ 2x^2 + 2y^2 - 40y + 150 = 0\\ x^2 + y^2 - 20y + 75 = 0$
Doug M
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@Almot1960 for any 2 vectors you have $(a,b)\cdot (c,d)=ac+bd=|(a,b)|\times|(c,d)|\cos\theta$ where $a,b,c,d$ are the $x,y$ values of the vector (for example a vector from the origin to the point $1,1$ will be $(1,1)$) the notation of $|\cdot|$ is the size of the vector, can be calculate by Pythagorean theorem ($|(x,y)|=\sqrt{x^2+y^2}$) and $\theta$ is the angle between the 2 vectors – ℋolo Dec 04 '17 at 22:27
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It is the Euclidean inner product. If you haven't learned that, skip to the next section where I derive the same equations using the Pythagorean theorem. – Doug M Dec 04 '17 at 22:28
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\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}
g.kov
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