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A. $ 2\sqrt 3 $

B. $ \sqrt 3 + \dfrac{{2\pi }}{3} $

C. $ 2\sqrt 3 + \dfrac{\pi }{3} $

D.None of these

Answer

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The given question when drawn on the sheet is shown below:

As given in the question the equation of the circle is $ {x^2} + {y^2} = 1 $ . Compare the equation of circle with general equation of circle $ {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} $ , where $ \left( {h,k} \right) $ is the centre of the circle and $ r $ is the radius of the circle.

After comparing it gives the centre of the circle as $ C\left( {0,0} \right) $ and the radius of the circle as $ 1 $ .

Draw the tangents from the points $ \left( { - 2,0} \right) $ and $ \left( {2,0} \right) $ to the circle $ {x^2} + {y^2} = 1 $ . It meets the circle at the points $ P\left( { - \dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}} \right) $ and $ Q\left( {\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}} \right) $ .

Now the triangles $ APC $ and $ BQC $ are both right angled triangles. So, by the help of Pythagoras theorem it can be deduced that $ AP = \sqrt {C{A^2} - C{P^2}} $ and $ BQ = \sqrt {C{B^2} - C{Q^2}} $ .

The lengths $ CA $ , $ CP $ , $ CB $ and $ CQ $ can be calculated with the help of distance formula and are equal to $ 2 $ , $ 1 $ , $ 2 $ and $ 1 $ respectively.

Substitute these values we get,

$

\Rightarrow AP = \sqrt {C{A^2} - C{P^2}} \\

= \sqrt {{2^2} - {1^2}} \\

= \sqrt {4 - 1} \\

= \sqrt 3 \\

$

And,

$

BQ = \sqrt {C{B^2} - C{Q^2}} \\

= \sqrt {{2^2} - {1^2}} \\

= \sqrt {4 - 1} \\

= \sqrt 3 \;

$

The arc length $ PQ $ will be equal to $ r \times \angle PCQ = 1 \times \dfrac{\pi }{3} = \dfrac{\pi }{3} $ .

The shortest distance from $ \left( { - 2,0} \right) $ to $ \left( {2,0} \right) $ is equal to sum of the lengths of $ AP $ , $ BQ $ and $ PQ $ i.e. $ \sqrt 3 + \sqrt 3 + \dfrac{\pi }{3} = 2\sqrt 3 + \dfrac{\pi }{3} $ .

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