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A circle centered at (3, 2) is tangent to the line y=(x+1)/3. Find the area of the circle.

 May 7, 2019
 #1
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(3,2)

The line can be written as    y = (1/3)(x + 1)     (1)

We need to find a perpendicular line to this one passing through (3,2)

This line will have a slope of  -3.....so we have

 

y = -3 ( x -3) + 2

y = -3x + 11        (2)

 

Set (1) and (2) equal to find the  x coordinate of the  point on the circle where these lines intersect

 

(1/3)(x + 1)  =  -3x + 11             multiply through by 3

x + 1  = -9x + 33      rearrange as

10x = 32

x = 32/10 =  16/5

And y  = -3(16/5) + 11  =   -48/5 + 55/5  =  7/5

 

So....the intersection is   (16/5, 7/5)

And the distance between this point   and (3,2)  will be the radius of the circle...so....we can find r^2  as

 

(3 - 16/5)^2 + (2 - 7/5)^2  =

(-1/5)^2 + (3/5)^2 =

1/25 + 9/25  =

10/25  =   2/5   = r^2

 

So....the area of the circle  =  pi * r^2  =    (2/5) pi units^2

 

 

Here's a graph : https://www.desmos.com/calculator/tklpk6vonx

 

 

cool cool cool

 May 8, 2019
 #2
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Thank you so much!!!

Guest May 8, 2019

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