Question

Two concentric circles having radii $73$ and $55$ are given. The chord of circle having a greater radius touches the smaller circle. Find the length of this chord.

Answer 1

Let $O$ be the centre of the concentric circles and $AB$ is the chord of the larger circle and touches the circle with the smaller radius at $M$.

$OA=$ radius of the greater circle $=73$

$OM=$ radius of the smaller circle $=55$

Also $OM\perp AB$ ........ ($AB$ is the tangent for smaller circle)

In $△OMA,\angle OMA={90}^o$

${OA}^2={OM}^2+{MA}^2$

$⟹{MA}^2={OA}^2-{OM}^2$

$={\left(73\right)}^2-{\left(55\right)}^2$

$=(73+55)(73-55)$

$=\left(128\right)\left(18\right)$

$={\left(48\right)}^2$

$\therefore$ $MA=48$

$\therefore$ $AB=2MA=2\left(48\right)=96$

Thus, the length of chord is $96$.