The pair of tangents drawn from an external point to a circle are perpendicular to each other and the length of each tangent (from the point to the circle) is 5 cm. The radius of the circle is

Let $P$ be the external point, $O$ be the center of the circle, $A$ and $B$ be the points where tangents touch the circle, and $r$ be the radius.

The length of each tangent is $PA = PB = 5$ cm and $\angle APB = 90°$.

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Tangents from an external point are perpendicular to the radius at the point of contact:

$PA \perp OA$ and $PB \perp OB$

Therefore $\angle PAO = \angle PBO = 90°$

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In quadrilateral $PAOB$, the sum of all angles equals $360°$:

$\angle PAO + \angle APB + \angle PBO + \angle AOB = 360°$

$90° + 90° + 90° + \angle AOB = 360°$

$\angle AOB = 90°$

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All angles in quadrilateral $PAOB$ are $90°$.

Since $PA = PB$ (tangent lengths are equal) and $OA = OB$ (radii are equal), the quadrilateral $PAOB$ is a square.

In a square, all sides are equal:

$PA = OA = r$

$r = 5$ cm

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Therefore, the radius of the circle is $5$ cm.