2023 AMC 12B Problems/Problem 10
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[hide]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 |
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All AMC 12 Problems and Solutions |
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