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(D) Let the circle have center $O$ and radius $r = 5 \, cm$. $AB$ is a diameter,so $OA = OB = 5 \, cm$.Tangent $XY$ is drawn at point $A$. Since the t

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$8$

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(D) Let the circle have center $O$ and radius $r = 5 \, cm$. $AB$ is a diameter,so $OA = OB = 5 \, cm$.Tangent $XY$ is drawn at point $A$. Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OA \perp XY$.Let the chord $CD$ be parallel to $XY$. Let $CD$ intersect the diameter $AB$ at point $E$. Since $CD \parallel XY$ and $OA \perp XY$,it follows that $OE \perp CD$.The distance of the chord $CD$ from $A$ is given as $AE = 8 \, cm$.Since $OA = 5 \, cm$,the distance of the chord from the center $O$ is $OE = AE - OA = 8 - 5 = 3 \, cm$.In the right-angled triangle $	riangle OEC$,by the Pythagorean theorem:$OC^2 = OE^2 + EC^2$$5^2 = 3^2 + EC^2$$25 = 9 + EC^2$$EC^2 = 16$$EC = 4 \, cm$.Since the perpendicular from the center to a chord bisects the chord,$CD = 2 	imes EC = 2 	imes 4 = 8 \, cm$.

At one end $A$ of a diameter $AB$ of a circle of radius $5 \, cm$,a tangent $XY$ is drawn to the circle. The length of the chord $CD$ parallel to $XY$ and at a distance of $8 \, cm$ from $A$ is (in $cm$):

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