Consider a circle whose radius is $5 m$. Let $AB$ be a chord in the circle of lenth $6 m$. And the two tangents at the points $A$ and $B$ are intersect at the point $P$.

In the following diagram, $CP=x$ and $AB=6$ and $OD=5$

Since $BL$ is an altitude of $\Delta OBP$ and $\measuredangle OBP=90^{\circ},$ we obtain
$$w^2=th$$ and
$$w^2=h\sqrt{r^2-w^2}.$$
Now, solve the last equation.

## Best Answer

It is easy to see, that $OP$ and $AB$ are perpendicual and intersects $AB$ in two equal parts, ie. $|AD|=|BD|=3$.

After Pythagoras: $$|AD|^2 + |OD|^2 = |OA|^2$$ $$3^2 + |OD|^2 = 5^2$$ we have then $|OD|=4$

Let $\alpha = âˆ DOA $ and $|OP|=y$ Then $$\cos \alpha = \frac{|OD|}{|OA|} = \frac{|OA|}{|OP|}$$ $$\frac{5}{y}= \frac{4}{5}$$

Then $y = \frac{25}{4}$, and $x=y-5=\frac{5}{4}$