Center of a circle from two chords.

Question

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.

Answer 1

The short answer is yes, it is possible.

The general outline of finding the distance $KO$ is this:

Given the lengths $AK, BK, CK$, and $DK$, use the Pythagorean Theorem to find the lengths of the sides of cyclic quadrilateral $ABCD$ and use Theorem 2 in this question to find the radius of the circle.

Once you have that, you can find the length of the perpendicular line from $AB$ passing through $O$. By Pythagoras, this is equal to $$m=\sqrt{r^2-\left(\frac{AK+BK}{2}\right)^2}$$ and, in turn, you can find $KO$ by using Pythagoras once again: $$KO=\sqrt{m^2+\left(\frac{AK-BK}{2}\right)^2}$$

Answer 2

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 $\dfrac{1}{2}\sqrt{(AB-2BK)^2+(CD-2CK)^2}$