Difference between revisions of "2023 AMC 12B Problems/Problem 10"
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+ | ==Problem== | ||
+ | In the <math>xy</math>-plane, a circle of radius <math>4</math> with center on the positive <math>x</math>-axis is tangent to the <math>y</math>-axis at the origin, and a circle with radius <math>10</math> with center on the positive <math>y</math>-axis is tangent to the <math>x</math>-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect? | ||
+ | <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 1== | ||
+ | |||
+ | The center of the first circle is <math>(4,0)</math>. | ||
+ | The center of the second circle is <math>(0,10)</math>. | ||
+ | Thus, the slope of the line that passes through these two centers is <math>- \frac{10}{4} = - \frac{5}{2}</math>. | ||
+ | |||
+ | 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 <math>\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}</math>. | ||
+ | |||
+ | |||
+ | ~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>. | ||
+ | [[Image:Amc12B_2023_P10.PNG|thumb|center|600px]] | ||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Video Solution 1 by OmegaLearn== | ||
+ | https://youtu.be/IUB6r1iNgpU | ||
+ | |||
+ | |||
+ | So um you can write both circles in polar form: The first circle with radius 4 um *long pause can be rewritten as r=8cosθ. And the other circle can be written as r=20sinθ. At the point of intersection, both radii will be equal. We can then do 8cosθ=20sinθ *clears throat. Rearrange the equation to solve for tanθ so tanθ = 2/5. Tanθ = y/x which yields the slope, so the slope is 2/5 <math>\boxed{\textbf{(E) } \frac{2}{5}}</math> moomoo. Thechickenonfire is the moomoo cow and you should milk the cow of knowledge. | ||
+ | |||
+ | |||
+ | -By Elite_Trash777 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/sjJMWtL_CEY | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2023|ab=B|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:54, 17 August 2024
Contents
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
.
Video Solution 1 by OmegaLearn
So um you can write both circles in polar form: The first circle with radius 4 um *long pause can be rewritten as r=8cosθ. And the other circle can be written as r=20sinθ. At the point of intersection, both radii will be equal. We can then do 8cosθ=20sinθ *clears throat. Rearrange the equation to solve for tanθ so tanθ = 2/5. Tanθ = y/x which yields the slope, so the slope is 2/5 moomoo. Thechickenonfire is the moomoo cow and you should milk the cow of knowledge.
-By Elite_Trash777
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.