If two chords of a circle intersect and are $\perp$ to each other, is it possible to find the distance from the intersection point of the chords to the center?
I was trying to use the power of a point argument.
[Math] Center of a circle from two chords.
circlesgeometry
Related Solutions
Another similar approach may be arisen as following: ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
The area of the upper red portion is
$$A_1=\frac{r^2}{2}\left(2\sin^{-1}\left(\frac{\sqrt{b^2+c^2-2bc\cos\phi}}{2r}\right) -\sin\left(2\sin^{-1}\left(\frac{\sqrt{b^2+c^2-2bc\cos\phi}}{2r}\right)\right)\right)+\frac{bc\sin\phi}{2}$$
This is due to the sum of the area of the triangle and segment bound by $b,c$ and the arc. The last term is the triangle and the first (larger) term is the segment based only on $r,b,c,\phi$ where $\phi$ is the angle between segments $b$ and $c$. Likewise, the are of the lower red portion is
$$A_2=\frac{r^2}{2}\left(2\sin^{-1}\left(\frac{\sqrt{a^2+d^2-2ad\cos\phi}}{2r}\right) -\sin\left(2\sin^{-1}\left(\frac{\sqrt{a^2+d^2-2ad\cos\phi}}{2r}\right)\right)\right)+\frac{ad\sin\phi}{2}$$
Same $\phi$ for both because of they are vertical angles. I'm sure that in conjunction with the identity $ab=cd$ that you started with, some kind of relationship between the two can be made.
Best Answer
Draw congruent chords parallel to those you have, and get this symmetric picture.
The central rectangle has width $AB-2BK$ and height $CD-2CK$. The center of the circle is at the intersection of the rectangle's diagonals, so the distance from $K$ to the center is half the length of a diagonal of the rectangle, or $\dfrac{1}{2}\sqrt{(AB-2BK)^2+(CD-2CK)^2}$