Math, asked by 99559955, 1 year ago

find the shortest distance from ( 0,0 ) to ( 12,12) without going inside the circle with centre ( 6,6 ) radius 5.

Answers

Answered by kvnmurty
0
See diagram.  

Let D be (12, 12)

Draw y = x line. Draw the circle  (x-6)^2 + (y-6)^2 = 5^2

The intersection points of these two are  A and B.

 Hence,     2 (x-6)^2 = 25
                x = 6 +  5 / √2  (point B)   or   6 - 5 / √2 (point A)
                y = x

We want the total of distance OA straight line distance, AEFB (semicircle perimeter, and straight line distance BD.

     OA = √2 * (6-5/√2)      
     AEFB = π * 5
     BD = √2 * (6-5/√2),     (OA and BD are equal due to symmetry.

Hence the Shortest distance = 12 √2 - 10 + 5 π

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Answered by dhiraj3
0
hence the shortest distance = 12√2-10+5π
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