A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm.

(c) 30 cm

Let AB be the chord of the given circle with centre O and a radius of 17 cm.
From O, draw OM perpendicular to AB.
Then OM = 8 cm and OB = 17 cm

From the right ΔOMB, we have:

$\mathrm{OB}^{2}=\mathrm{OM}^{2}+\mathrm{MB}^{2}$

$\Rightarrow 17^{2}=8^{2}+\mathrm{MB}^{2}$

$\Rightarrow 289=64+\mathrm{MB}^{2}$

$\Rightarrow \mathrm{MB}^{2}=(289-64)=225$

$\Rightarrow \mathrm{MB}=\sqrt{225} \mathrm{~cm}=15 \mathrm{~cm}$

The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AB = 2 × MB = (2 x 15) cm = 30 cm
Hence, the required length of the chord is 30 cm.