Question

From a point on the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 } \sin ^ { 2 } \alpha$ The angle between them is

• A. α
• B. $\frac { \alpha } { 2 }$
• C. $2 \alpha$
• D. None of these

Answer 1

Answer: (C)

$2 \alpha$

Answer 2

Let any point on the circle $( a \cos t , a \sin t )$ and $\angle O P Q = \theta$

Now; PQ = length of tangent from P on the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 } \sin ^ { 2 } \alpha$

$P Q = \sqrt { a ^ { 2 } \cos ^ { 2 } t + a ^ { 2 } \sin ^ { 2 } t - a ^ { 2 } \sin ^ { 2 } \alpha }$ $= a \cos \alpha$

$x ^ { 2 } + y ^ { 2 } = a ^ { 2 } \sin ^ { 2 } \alpha$ $\tan \theta = \frac { O Q } { P Q } = \tan \alpha$

$\Rightarrow \theta = \alpha$;

$\therefore$ Angle between tangents

Alternative Method : We know that, angle between the tangent from $( \alpha , \beta )$ to the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ is $2 \tan ^ { - 1 } \left( \frac { a } { \sqrt { \alpha ^ { 2 } + \beta ^ { 2 } - a ^ { 2 } } } \right)$.

Let point on the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ be $( a \cos t , a \sin t )$

Angle between tangent

= $2 \tan ^ { - 1 } \left( \frac { a \sin \alpha } { \sqrt { a ^ { 2 } \cos ^ { 2 } t + a ^ { 2 } \sin ^ { 2 } t - a ^ { 2 } \sin ^ { 2 } \alpha } } \right)$

$= 2 \alpha$