Difference between revisions of "2017 AMC 12B Problems/Problem 6"

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==Solution==
 
==Solution==
 
Solution 1  
 
Solution 1  
Because the two points are a diameter the center must be half way between them at the point (4,3). The distance from (0,0) to (4,3) is 5 therefore the circle has a radius of 5. The equation of the circle can thus be written as (x-4)^2+(y-3)^2=25. To find the x intercept y must be 0, so (x-4)^2+(0-3)^2=25. Then (x-4)^2=16. x-4=4. x=8
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Because the two points are a diameter the center must be half way between them at the point (4,3). The distance from (0,0) to (4,3) is 5 therefore the circle has a radius of 5. The equation of the circle can thus be written as (x-4)^2+(y-3)^2=25. To find the x intercept y must be 0, so (x-4)^2+(0-3)^2=25.  
 +
Then (x-4)^2=16.  
 +
x-4=4.  
 +
x=8
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=5|num-a=7}}
 
{{AMC12 box|year=2017|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:34, 16 February 2017

Problem 6

The circle having $(0,0)$ and $(8,6)$ as the endpoints of a diameter intersects the $x$-axis at a second point. What is the $x$-coordinate of this point?

Solution

Solution 1 Because the two points are a diameter the center must be half way between them at the point (4,3). The distance from (0,0) to (4,3) is 5 therefore the circle has a radius of 5. The equation of the circle can thus be written as (x-4)^2+(y-3)^2=25. To find the x intercept y must be 0, so (x-4)^2+(0-3)^2=25. Then (x-4)^2=16. x-4=4. x=8

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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