Question

in the figure below, the segments $overline{uv}$ and $overline{uw}$ are tangent to the circle centered at $o$. given that $ov = 3$ and $ou = 5$, find $uw$.

Explanation:

Step1: Recall tangent - radius property

A tangent to a circle is perpendicular to the radius at the point of tangency. So, \(\angle OVU = 90^{\circ}\) and \(\angle OWU=90^{\circ}\). Also, tangents from a common external point to a circle are equal in length, so \(UV = UW\).

Step2: Apply Pythagorean theorem in \(\triangle OVU\)

In right - triangle \(OVU\), we know that \(OU = 5\) (hypotenuse) and \(OV = 3\) (one leg), and we want to find the length of the other leg \(UV\). By the Pythagorean theorem, for a right - triangle with legs \(a\), \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\). Let \(UV=x\), then \(x^{2}+3^{2}=5^{2}\).
\[x^{2}=5^{2}-3^{2}\]
\[x^{2}=25 - 9\]
\[x^{2}=16\]
Taking the square root of both sides (and since length is positive), we get \(x = 4\). Since \(UV = UW\) (tangents from \(U\) to the circle), \(UW=4\).

Answer:

\(4\)