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avatar+2498 

If a circle is centered at the origin and tangent to the line y=-x+4 , what is the area of the circle devided by pi ?

A) 2

B) 4 

C) 8

D) 12

E) 16

 Dec 26, 2015

Best Answer 

 #2
avatar+128407 
+10

Notice that a tangent line drawn from the center of the circle to the point of tangency will meet the given line at right angles..........thus, these two lines will have reciprocal slopes....

 

And the equation of the line through the origin will be

 

y = x

 

So......setting these two equations equal, we have

 

x = -x + 4      add x to both sides

 

2x = 4

 

x = 2

 

So, at the point of tangency, y = x, so  y = 2

 

And the radius of such a circle centered at the origin will be

 

sqrt (2^2 + 2^2) = sqrt (8)

 

So....the area of this circle will be

 

A = pi [sqrt(8)]^2   =  8pi     and dividing this by pi   will produce a result of 8

 

Here's a graph of the situation ......https://www.desmos.com/calculator/g8eurlqsyo

 

 

cool cool cool

 Dec 26, 2015
edited by CPhill  Dec 26, 2015
 #1
avatar+2498 
0
 Dec 26, 2015
edited by Solveit  Dec 26, 2015
edited by Solveit  Dec 26, 2015
edited by Solveit  Dec 26, 2015
edited by Solveit  Dec 26, 2015
 #2
avatar+128407 
+10
Best Answer

Notice that a tangent line drawn from the center of the circle to the point of tangency will meet the given line at right angles..........thus, these two lines will have reciprocal slopes....

 

And the equation of the line through the origin will be

 

y = x

 

So......setting these two equations equal, we have

 

x = -x + 4      add x to both sides

 

2x = 4

 

x = 2

 

So, at the point of tangency, y = x, so  y = 2

 

And the radius of such a circle centered at the origin will be

 

sqrt (2^2 + 2^2) = sqrt (8)

 

So....the area of this circle will be

 

A = pi [sqrt(8)]^2   =  8pi     and dividing this by pi   will produce a result of 8

 

Here's a graph of the situation ......https://www.desmos.com/calculator/g8eurlqsyo

 

 

cool cool cool

CPhill Dec 26, 2015
edited by CPhill  Dec 26, 2015
 #3
avatar+2498 
0

thanks CPhill ! :)

 Dec 26, 2015

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