Question

Find the radius of the outer circle when chord of the outer circle which is the tangent to the inner circle measures $8cm$ and the radius of the inner circle is $3cm$.

• A. $11$ cm
• B. $5$ cm
• C. $4$ cm
• D. $6$ cm

Answer 1

The correct option is B $5$ cm

Here, radius of inner circle $OM=3$

Radius of outer circle $OB=r_2$

length of chord $AB=8cm$.

We need to find $r_2$, [$r_2$ is radius of outer circle]

Since the point of tangency of innier circle is the mid-point of the chord to outer circle;

M bisects AB.

Hence, $AM=MB=\frac{1}{2}.AB$.

Also we know that the radius of a circle is perpendicular to the tangent at the point of contact,

thus, $\angle OMB={90}^{\circ }$.

This implies,

triangle OMB is a right triangle and so by Pythagoras theorem we get,

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

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

$OB=\sqrt{3^2+4^2}$

$OB=\sqrt{25}$

$OB=5cm$

Hence Radius of outer circle is 5cm

Option B is the answer.