Distance From Center Of Circle To Intersection Of Diagonals Of A Cyclic Quadrilateral Given a cyclic quadrilateral $ABCD$ with length $AB=w$, $BC=x$, $CD=y$, and $DA=z$, compute the distance from the center of the circumscribing circle to the intersection point of the diagonals $AC$ and $BD$ in terms of $w,x,y,z$.
 A: I will use picture and the notations from
https://en.wikipedia.org/wiki/Cyclic_quadrilateral
so the cyclic quadrilateral under study is $ABCD$, it sides are $AB=a$, $BC=b$, $CD=c$, $DA=d$ (instead of $w,x,y,z$, which i cannot type in this context without error.)
The intersection of the diagonals is $P$, and the center of the circle $(ABCD)$ is $O$, we denote by $R$ the radius $R=OA=OB=OC=OD$.
When two letters are used in connection with metric relation, we mean always the corresponding lengths.  

We will use the following facts and solve the issue:



*

*The power of the interior point $P$ is 
$$PA\cdot PC=PB\cdot PD=R^2-OP^2\ .$$

*The relations of Brahmagupta and Parameshvara, and similar relations for the trigonometric functions of the angles in $A,B,C,D$, and of the angle $\theta$ between the diagonals.

*The sine theorem, for instance 
$$
\begin{aligned}
\frac{PA}{DA}
&=
\frac
{\sin \widehat{PDA}}
{\sin \widehat{APD}}
=
\frac
{\sin \widehat{BDA}}
{\sin \theta}
=
\frac
{\sin \frac 12\widehat{BOA}}
{\sin \theta}
=
\frac {\frac{AB/2}R}
{\sin \theta}\ ,
\\[2mm]
&\qquad
\text{ which implies}
\\[2mm]
PA 
&= \frac{AD\cdot AB}{2R\sin\theta}
= \frac{da}{2R\sin\theta}
\ ,\qquad\text{ and similarly}
\\
PC &= \frac{CD\cdot CB}{2R\sin\theta}
= \frac{cb}{2R\sin\theta}
\ .
\end{aligned}
$$

We are now in position to join the following relations:
$$
\begin{aligned}
R^2-OP^2
&=
PA\cdot PC
=\frac {abcd}{4R^2\sin^2\theta}\ ,
\\
4R^2 
&=
\frac 14\cdot
\frac
{(ab+cd)(ac+bd)(ad+bc)}
{(s-a)(s-b)(s-c)(s-d)}\text{ (Parameshvara) with }
\\
s&=\frac 12(a+b+c+d)\ ,
\\[2mm]
&\qquad\text{ and we use now}
\\
\tan^2\frac\theta 2
&=
\frac{(s-b)(s-d)}{(s-a)(s-c)}\text{ to compute}
\\
\frac1{\cos^2\frac\theta 2}
&=
1+
\tan^2\frac\theta 2
=
\frac{(s-a)(s-c)+(s-b)(s-d)}{(s-a)(s-c)}
\sin^2\theta
\\
\cos^2\frac\theta 2
&=
\frac
{(s-a)(s-c)}
{(s-a)(s-c)+(s-b)(s-d)}
\\
\sin^2\frac\theta 2
&=
\frac
{(s-b)(s-d)}
{(s-a)(s-c)+(s-b)(s-d)}
\\
\sin^2\theta &= 
4
\sin^2\frac\theta 2
\cos^2\frac\theta 2
\\
&=
4\cdot
\frac
{(s-a)(s-b)(s-c)(s-d)}
{(\; (s-a)(s-c)+(s-b)(s-d)\;)^2}\ ,
\\
4R^2\sin^2\theta
&=
\frac
{(ab+cd)(ac+bd)(ad+bc)}
{(\;(s-a)(s-c)+(s-b)(s-d)\;)^2}\ ,
\\
PA\cdot PC
&=\frac {abcd}{4R^2\sin^2\theta}
=
\frac
{abcd\;(\;(s-a)(s-c)+(s-b)(s-d)\;)^2}
{(ab+cd)(ac+bd)(ad+bc)}
\ ,\\[2mm]
&\qquad\text{ and finally}
\\
\color{blue}
{OP^2}
&=
R^2-PA\cdot PC
\\
&=
\color{blue}
{
\frac 1{16}
\cdot
\frac 1{(s-a)(s-b)(s-c)(s-d)}
\cdot
\frac
{(ac+bd)^2}
{(ab+cd)(ad+bc)}
\cdot
\Big(\ 
bd(a^2-c^2)^2+ac(b^2-d^2)^2
\ \Big)
}\ .
\end{aligned}
$$

Note:
At the last step we have used sage to factorize. Code and results:
sage: S.<a,b,c,d> = PolynomialRing(QQ)
sage: RR = 1/16 * (a*c+b*d)*(a*b+c*d)*(a*d+b*c) / (s-a) / (s-b) / (s-c) / (s-d)
sage: PAPC = a*b*c*d * ((s-a)*(s-c)+(s-b)*(s-d))^2 / (  (a*c+b*d)*(a*b+c*d)*(a*d+b*c) )
sage: factor( RR - PAPC )
-(a*b^4*c + a^4*b*d - 2*a^2*b*c^2*d + b*c^4*d 
                    - 2*a*b^2*c*d^2 + a*c*d^4)
* (a*c + b*d)^2
/ ((a*b + c*d)*(b*c + a*d)
  *(a + b + c - d)
  *(a + b - c + d)
  *(a - b + c + d)
  *(a - b - c - d))

(Lines were manually broken.)
A: As in @dan_fulea's answer, I'll take the sides of the cyclic quadrilateral $\square ABCD$ to be $a$, $b$, $c$, $d$ (in order). Let the diagonals be $p$ and $q$ (either order). Let the diagonals meet at $L$, let their midpoints be $M$ and $N$ (either order), and let the circumcenter be $O$.
Each diagonal is a chord of the circle; therefore, it's perpendicular to the segment joining its midpoint to $O$. Consequently, $\square OMLN$ has an opposite pair of right angles, making it cyclic; moreover, its circumdiameter is exactly $|OL|$, the distance we seek. By the (Extended) Law of Sines and the area formula $|\square ABCD|=\tfrac12pq\sin\angle MLN$ (valid for any quadrilateral), we can write

$$|OL| = \frac{|MN|}{\sin\angle MLN} = \frac{p q \;|MN|}{2\;|\square ABCD|} \tag{$\star$}$$

From here, we proceed as @dan did, leveraging existing results to get expressions for substitution into $(\star)$.

Euler's (not-necessarily-cyclic) Quadrilateral Theorem states that
$$a^2 + b^2 + c^2 + d^2 = p^2 + q^2 + 4 |MN|^2 \tag{1}$$
and it is "known" that the diagonals of a cyclic quadrilateral satisfy
$$p^2 + q^2 = \frac{(ac+bd)(ad+bc)}{ab+cd}+\frac{(ac+bd)(ab+cd)}{ad+bc} \qquad\qquad
pq = ac+bd \tag{2}$$
(the latter of which is Ptolemy's Theorem). Finally, Brahmagupta's formula for the area of a cyclic quadrilateral gives
$$|\square ABCD|^2 = (s-a)(s-b)(s-c)(s-d) \tag{3}$$
where $s:=\tfrac12(a+b+c+d)$ is the semiperimeter of $\square ABCD$.
Substituting $p^2+q^2$ from $(2)$ into $(1)$, and solving, we get
$$|MN|^2 = \frac{ac(b^2-d^2)^2+bd(a^2-c^2)^2}{4(ab+cd)(ad+bc)} \tag{4}$$
Therefore, squaring to avoid radicals, $(\star)$ becomes ...

$$|OL|^2 = \frac{1}{16}\;\frac{ac(b^2-d^2)^2+bd(a^2-c^2)^2}{(s-a)(s-b)(s-c)(s-d)}\;\frac{(ac+bd)^2}{(ab+cd)(ad+bc)} \tag{$\star\star$}$$

... which agrees with @dan's solution. $\square$
