What is the length of line X and line Y such that, together with line B, they form a right triangle?
Note: -the vertex between line X and line Y should be a point in the circle -the angle between line X and line Y is 90 degrees
A rephrasing of the problem would be: "Given the right triangle XYB tangent to the circle at vertex B, and line A intersecting the circle's center and bisecting side B, what is the length of the sides X and Y such that side B is perpendicular to line A?"
Given $|OT|=|OB|=r=26$, $XY\perp OT$, $|XY|=b=5.374295$ $|MX|=|MY|=\tfrac12b$, $|OM|=d=23.849093$, $\angle XBY=90^\circ$.
Find $|BX|,|BY|$.
Let $O=(0,0)$, $T=(0,r)$. Then \begin{align} X=( \tfrac12b,d)=(&2.6871475,23.849093) ,\\ Y=(-\tfrac12b,d)=(-&2.6871475,23.849093) , \end{align}
Note that since $\angle XBY=90^\circ$, $M$ is the center of circumscribed circle around $\triangle XBC$, we have
\begin{align} |MB|&=|MX|=|MY|=\tfrac12b, \\ \cos\phi&=\frac{r^2+d^2-(\tfrac12b)^2}{2\,r\,d} =0.997908031422582 ,\\ B&=(r\,\sqrt{1-\cos^2\phi},\ r\,\cos\phi) =(1.68088759763376,25.9456088169871) ,\\ |BX|&=\sqrt{(B_x-X_x)^2+(B_y-X_y)^2} =2.32549727198021 ,\\ |BY|&=\sqrt{(B_x-Y_x)^2+(B_y-Y_y)^2} =4.8451118857089 . \end{align}