Question

At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. Find the length of the chord CD parallel to XY and at a distance 8 cm from A.

Answer 1

A chord $CD$ is drawn which is parallel to $XY$ and at a distance of $8cm$ from $A$.

As we know that tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore \angle OAY=90°$

As sum of cointerior angle is $180°$.

Therefore,

$\angle OAY+\angle OED=180°$

$\Rightarrow \angle OED=90°$

$AE=8cm\left(\text{From fig.}\right)$

Now in $△OEC$, by pythagoras theorem,

${OC}^2={OE}^2+{EC}^2$

$\Rightarrow {EC}^2={OC}^2-{OE}^2$

$\Rightarrow {EC}^2={\left(5\right)}^2-{\left(3\right)}^2$

$\Rightarrow EC=\sqrt{25-9}=4$

Therefore,

Length of chord $CD=2\times CE(\because \text{perpendicular from centre to the chord bisects the chord})$

$\Rightarrow CD=2\times 4=8cm$

Hence the length of the chord $CD$ is $8cm$.