Difference between revisions of "2023 AMC 12B Problems/Problem 10"
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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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− | 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. | + | 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. |
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.