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**Find the length of the shortest path from (0,0) to (12,16) in the x-y plane which does not go inside the curve x^2-12x+y^2-16y+75=0**

Written by Implex

September 11, 2008 at 1:51 pm

Posted in Coordinate Geometry, Problem of the week

Tagged with Coordinate Geometry, Problem of the week

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I’ve found out that the curve is a closed figure and . Now, we need to find the lines that join (0,0) and (12,16) and are tangent to the curve. Not able to move further. help!!

We can find the equations of the lines that pass through (0,0) and (1,8), and (12,16) and the topmost point of the figure. Their point of intersection will give us the co-ordinates of the point.Then finding distance is simple.

milindSeptember 11, 2008 at 4:59 pm

hmmm, you were on right track till, you did some mistake !

outtimedSeptember 11, 2008 at 5:18 pm

The equation is a circle with center at (6,8) and radius 5. So the only possible shortest distance satisfying the given condition must be sum of lengths of two tangents (one from [0,0] and other from [12,16]) and then extending the tangent to intersect each other.

radius = 5 and distance from origin to center = 10. Therefore length of Tangent PT^2 = 10*(10-5) => PT = 5(2)^1/2.

The 4 tangents when joined results in a parallelogram circumscribing the given circle. Also one of the angles of the parallelogram would be 60 as Perpendicular = 5, Hypotenuse=10. The other would be 120.

Distance = 10{3*sqrt(2)+sqrt(3)}/3 = 19.91 Approx. Ans.

TanujSeptember 12, 2008 at 3:27 am

10+5pi .. right ??

AnshulSeptember 17, 2008 at 6:06 am

what is the correct answer outtimed and wats the best method?

CATastrophe08October 2, 2008 at 10:31 pm

the given curve is a circle with center at (6,8) and radius = 5.

is the shortest path = 17 + 5*pi/2 .

yodhaOctober 10, 2008 at 3:23 pm