91Ó°ÊÓ

Given:

The circle with a center $O$with perpendicular segment $OX$and segment $OY$.

The tangent meet at $Z$.

Now, figure drawn by the information is,

In quadrilateral $OXYZ,$$OXâŠ\yen XZ\text{and}OYâŠ\yen YZ.$

$OX$ and $OY$ is tangent to a circle.

That is;

$\begin{array}{l}OX\left|\right|YZ\\ OY\left|\right|XZ\end{array}$ .......(1)

Then addition of Adjutants angle is $180°$.

$\begin{array}{l}∠X+∠Y=180°\\ \begin{array}{l}90°+∠Y=180°\\ ∠Y=90°\end{array}\end{array}$

Similarly prove that, $∠Z=90°$.

Then all angles of quadrilateral are $90°$ and opposite side are parallel.

Therefore, this quadrilateral is Rectangle.

Thus,

$\begin{array}{l}OX=YZ\\ OY=XZ\mathrm{......}\left(2\right)\end{array}$

But $YX=XY$ (Tangent to the circle to same point)

Then, it can be concluded that,

$\begin{array}{l}OX=OY\\ =YZ\\ =XY\end{array}$

Then, all side is equal and all angles are $90°$.

The given value is,

$OX=5$

In the above figure this is a square then all side is same.

In triangle $OXZ,$

$\begin{array}{l}\sqrt{O{Z}^2}=\sqrt{50}\\ OZ=5\sqrt{2}\end{array}$

Therefore, the value of $OZ$is, $OZ=5\sqrt{2}$.