Difference between revisions of "2023 AMC 12B Problems/Problem 10"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
This is in slope intercept form therefore the slope is <math>\boxed{\textbf{(E) } \frac{2}{5}}</math>. | This is in slope intercept form therefore the slope is <math>\boxed{\textbf{(E) } \frac{2}{5}}</math>. | ||
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| + | ~luckuso | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2023|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2023|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:45, 15 November 2023
Problem
In the 
-plane, a circle of radius 
with center on the positive 
-axis is tangent to the 
-axis at the origin, and a circle with radius 
with center on the positive -axis is tangent to the -axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?
Solution 1
The center of the first circle is 
.
The center of the second circle is 
.
Thus, the slope of the line that passes through these two centers is 
.
Because this line is the perpendicular bisector of the line that passes through two intersecting points of two circles, the slope of the latter line is 
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Coordinate Geometry)
The first circle can be written as 
we'll call this equation 
The second can we writen as 
, we'll call this equation 
Expanding :
Exapnding
Now we can set the equations equal to eachother:
This is in slope intercept form therefore the slope is 
.
~luckuso
See Also
| 2023 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
