2023 AMC 12B Problems/Problem 10: Difference between revisions
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<math>\textbf{(A)}\ \dfrac{2}{7} \qquad\textbf{(B)}\ \dfrac{3}{7} \qquad\textbf{(C)}\ \dfrac{2}{\sqrt{29}} \qquad\textbf{(D)}\ \dfrac{1}{\sqrt{29}} \qquad\textbf{(E)}\ \dfrac{2}{5}</math> | <math>\textbf{(A)}\ \dfrac{2}{7} \qquad\textbf{(B)}\ \dfrac{3}{7} \qquad\textbf{(C)}\ \dfrac{2}{\sqrt{29}} \qquad\textbf{(D)}\ \dfrac{1}{\sqrt{29}} \qquad\textbf{(E)}\ \dfrac{2}{5}</math> | ||
==Solution== | ==Solution 1== | ||
The center of the first circle is <math>(4,0)</math>. | The center of the first circle is <math>(4,0)</math>. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==Solution 2 (Coordinate Geometry)== | |||
The first circle can be written as <math>(x-4)^2 + y^2 = 4^2</math> we'll call this equation <math>(1)</math> | |||
The second can we writen as <math>x^2 + (y-10)^2 = 10^2</math>, we'll call this equation <math>(2)</math> | |||
Expanding <math>(1)</math>: | |||
<cmath>\begin{align*} | |||
x^2 -8x + 4^2 + y^2 &= 4^2 \\ | |||
x^2 - 8x + y^2 &= 0 | |||
\end{align*}</cmath> | |||
Exapnding <math>(2)</math> | |||
<cmath>\begin{align*} | |||
x^2 + y^2 -20y + 10^2 = 10^2\\ | |||
x^2 + y^2 - 20y = 0 | |||
\end{align*}</cmath> | |||
Now we can set the equations equal to eachother: | |||
<cmath>\begin{align*} | |||
x^2 - 8x + y^2 &= x^2 + y^2 - 20y \\ | |||
\frac{8}{20}x &= y \\ | |||
\frac{2}{5}x &= y | |||
\end{align*}</cmath> | |||
This is in slope intercept form therefore the slope is <math>\boxed{\textbf{(E) } \frac{2}{5}}</math>. | |||
==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 21:44, 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 
.
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 | |
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