Question

In the given figure, there are two concentric circles with center O such that AP is tangent to the bigger circle and AB is tangent to the smaller circle. If $\angle APB=\angle ABP={30}^{\circ },OA=3$ cm and OP =5 cm, then, radius of the smaller circle is

• A. 5 cm
• B. $\sqrt{5}cm$
• C. 6 cm
• D. $\sqrt{6}cm$

Answer 1

The correct option is B

$\sqrt{5}cm$

Given that, there are two concentric circles with center O. PA is tangent to bigger circle and AB is tangent to the smaller circle.

$\Rightarrow OA\perp APandOR\perp AB$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)
In $\Delta OAP,$
$O{A}^2+A{P}^2=O{P}^2\left(Pythagorastheorem\right)\Rightarrow 3^2+A{P}^2=5^2\Rightarrow A{P}^2=16\Rightarrow AP=4cm$
$\angle APB=\angle ABP={30}^{\circ }\left(Given\right)\Rightarrow AP=AB=4cm\left(sidesoppositetoequalanglesareequal\right)\Rightarrow AB=4cm$
Now, AB is chord to bigger circle with $OR\perp AB.$
So, OR bisects AB.
[perpendicular from the centre to the chord, bisects the chord]
$\Rightarrow AR=RB=2cm$
Now, In $\Delta ORA,$
$O{A}^2=O{R}^2+A{R}^2\Rightarrow 3^2=O{R}^2+2^2\Rightarrow 9-4=O{R}^2\therefore OR=\sqrt{5}cm$
So, the radius of the smaller circle is $\sqrt{5}cm$.