The perimeter of the quadrilateral $APOQ$ is $26\ cm$ and $AP = 10\ cm$. $O$ is the centre of the circle. Then the radius of the circle is (where $AP$ and $AQ$ are the tangents from $A$)

Because AP and AQ are tangents from external point A, their lengths are equal, so AQ is also 10 cm. Let the circle radius be r, making the lengths OP and OQ equal to r. The perimeter of quadrilateral APOQ is AP plus AQ plus OP plus OQ, which gives the equation 10 plus 10 plus r plus r equals 26. Solving this equation, 20 plus 2r equals 26, meaning the radius r is 3 cm.

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