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find the exact length of the tangent from (4,-5) to the circle x^2+4x+y^2-2y-11=0  ...

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bubbletrea | Student, Grade 11 | Honors

Posted July 8, 2013 at 2:44 AM via web

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find the exact length of the tangent from (4,-5) to the circle x^2+4x+y^2-2y-11=0

 

ans =56^1/2

Tagged with circle as a locus, math

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aruv | High School Teacher | Valedictorian

Posted July 8, 2013 at 2:58 AM (Answer #1)

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tangent from (4,-5) to the circle x^2+4x+y^2-2y-11=0

The equation of circle can be written as

`(x+2)^2+(y-1)^2=11+4+1`

`(x+2)^2+(y-1)^2=(4)^2`

So centre of the circle at (-2,1) and radius r of the circle is 4

The distance d of the centre from point (4,-5) is

`d^2=(4+2)^2+(-5-1)^2`

`=36+36=72`

`` Thus length L of the tangents from (4,-5) is

`d^2=L^2+r^2`

`72=L^2+16`

`L^2=72-16=56`

`L=sqrt(56)`

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